Exponents

Study Sheet

Exponents

Every rule, tip, and worked example in one place

What an Exponent Means

Concept
Repeated Multiplication

An exponent is a shortcut for repeated multiplication. In the power ana^{n}:

  • aa is the base — the number being multiplied.
  • nn is the exponent (or power) — how many times the base is used as a factor.
an=a⋅a⋅a⋯a⏟n factors,25=2⋅2⋅2⋅2⋅2=32.a^{n} = \underbrace{a \cdot a \cdot a \cdots a}_{n \text{ factors}}, \qquad 2^{5} = 2\cdot 2\cdot 2\cdot 2\cdot 2 = 32.

We read ana^{n} as “aa to the nnth power.” Special names: a2a^{2} is “aa squared” and a3a^{3} is “aa cubed.”

The exponent rules let you compress or expand products of like bases; zero and negative exponents fill in the pattern; scientific notation uses powers of ten to tame huge and tiny numbers. Exponential growth (y=a⋅bxy = a\cdot b^x with b>1b > 1) multiplies by the same factor each step, which is why it eventually outruns any linear growth.

Every exponent rule in this topic is a way of counting factors, so the definition is the whole theory. Keep the picture "ana^n is nn copies of aa multiplied" and each rule can be rebuilt on the spot instead of memorized; the zero and negative exponents are then defined so that the counting rules keep working.

Example
Expanding and evaluating

Write each power as repeated multiplication, then evaluate.

34=3⋅3⋅3⋅3=81,(−2)3=(−2)(−2)(−2)=−8,(12)3=12⋅12⋅12=18.\begin{aligned} 3^{4} &= 3\cdot 3\cdot 3\cdot 3 = 81, \\[2pt] (-2)^{3} &= (-2)(-2)(-2) = -8, \\[2pt] \left(\tfrac{1}{2}\right)^{3} &= \tfrac{1}{2}\cdot\tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{8}. \end{aligned}
Side note
Where it goes wrong

Treating 2−32^{-3} as negative. Negative exponents mean reciprocals: 2−3=18>02^{-3} = \tfrac18 > 0. The sign of an exponent controls where the factors go (numerator or denominator), never the sign of the result.

Tip

Watch the sign carefully: (−2)4=16(-2)^{4} = 16 (four negatives multiply to a positive), but −24=−(24)=−16-2^{4} = -(2^{4}) = -16, because without parentheses the exponent applies only to the 22, and the minus sign waits out front.

Try it
Try it: evaluate carefully

Evaluate (−3)4(-3)^4, −34-3^4, (23)3\left(\tfrac23\right)^3 and 0.130.1^3.

Show the answer

(−3)4=81(-3)^4 = 81 (four negative factors). −34=−(34)=−81-3^4 = -(3^4) = -81. (23)3=827\left(\tfrac23\right)^3 = \tfrac{8}{27}. 0.13=0.0010.1^3 = 0.001.

All the Exponent Rules at a Glance

Concept
The Complete Rule Sheet

For any nonzero base(s) aa, bb and integers mm, nn:

  • Product of powers: am⋅an=am+na^{m}\cdot a^{n} = a^{m+n}
  • Quotient of powers: aman=am−n\dfrac{a^{m}}{a^{n}} = a^{m-n}
  • Power of a power: (am)n=amn\left(a^{m}\right)^{n} = a^{mn}
  • Power of a product: (ab)n=anbn\left(ab\right)^{n} = a^{n}b^{n}
  • Power of a quotient: (ab)n=anbn\left(\dfrac{a}{b}\right)^{n} = \dfrac{a^{n}}{b^{n}}
  • Zero exponent: a0=1a^{0} = 1
  • Negative exponent: a−n=1ana^{-n} = \dfrac{1}{a^{n}}   and   1a−n=an\dfrac{1}{a^{-n}} = a^{n}

The whole topic comes back to one idea: same base, combine the exponents.

All seven rules come from counting factors of the base. Multiplying strings together adds the counts, dividing cancels factors and subtracts the counts, and stacking powers multiplies the counts. The zero and negative rules are what those counting rules force when the counts run out.

Side note
Where it goes wrong

Mixing up which rule applies: (x3)4=x7(x^3)^4 = x^7 or x3⋅x4=x12x^3 \cdot x^4 = x^{12}. Stacking powers multiplies (x12x^{12}); multiplying powers adds (x7x^7). When unsure, write out a small case: (x2)3=x2⋅x2⋅x2=x6(x^2)^3 = x^2 \cdot x^2 \cdot x^2 = x^6.

Try it
Try it: one line each

Simplify a4⋅a5a^4 \cdot a^5, b9b3\dfrac{b^9}{b^3}, (c2)5(c^2)^5, (2d)3(2d)^3 and e−2⋅e2e^{-2} \cdot e^{2}.

Show the answer

a9a^9; b6b^6; c10c^{10}; 8d38d^3 (the 22 is cubed too); e0=1e^0 = 1.

Product of Powers

Concept
Add the Exponents

When you multiply powers with the same base, keep the base and add the exponents.

am⋅an=am+n.a^{m}\cdot a^{n} = a^{m+n}.

Why? a2⋅a3=(a⋅a)(a⋅a⋅a)=a5a^{2}\cdot a^{3} = (a\cdot a)(a\cdot a\cdot a) = a^{5}: you are just counting all the factors.

Writing a2⋅a3a^2 \cdot a^3 out in full gives two factors of aa followed by three more, five in all. Nothing about the base mattered, only the counting, which is why the rule holds for any base and why the bases must match: x2⋅y3x^2 \cdot y^3 is two xx's and three yy's, which cannot be counted together.

Example
Multiplying like bases
aa·aaaa²a³aaaaa2 + 3 = 5 factors: a²·a³ = a⁵
Two factors and three factors make five.
x4⋅x3=x4+3=x7,25⋅22=27=128,5a2⋅3a6=(5⋅3) a2+6=15a8.\begin{aligned} x^{4}\cdot x^{3} &= x^{4+3} = x^{7}, \\[2pt] 2^{5}\cdot 2^{2} &= 2^{7} = 128, \\[2pt] 5a^{2}\cdot 3a^{6} &= (5\cdot 3)\,a^{2+6} = 15a^{8}. \end{aligned}

Multiply the coefficients (the plain numbers) and add the exponents of the matching variables.

Side note
Where it goes wrong

Multiplying the exponents: x2⋅x3=x6x^2 \cdot x^3 = x^6. Count them: (x⋅x)(x⋅x⋅x)(x \cdot x)(x \cdot x \cdot x) has five factors, so x5x^5. Test with x=2x = 2: 4⋅8=32=254 \cdot 8 = 32 = 2^5, not 64=2664 = 2^6.

Tip

The bases must match. You cannot combine x3⋅y2x^{3}\cdot y^{2} into a single power — different bases stay separate.

Try it
Try it: product rule

Simplify (3x2y)(−4x5y3)(3x^2y)(-4x^5y^3).

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)
Show the answer

Multiply the coefficients: 3⋅(−4)=−123 \cdot (-4) = -12. Add exponents on each base: x2+5=x7x^{2 + 5} = x^7 and y1+3=y4y^{1 + 3} = y^4. Result: −12x7y4-12x^7y^4.

Quotient of Powers

Concept
Subtract the Exponents

When you divide powers with the same base, keep the base and subtract the bottom exponent from the top.

aman=am−n.\frac{a^{m}}{a^{n}} = a^{m-n}.

The shared factors cancel: a5a2=a⋅a⋅a⋅a⋅aa⋅a=a3\dfrac{a^{5}}{a^{2}} = \dfrac{a\cdot a\cdot a\cdot a\cdot a}{a\cdot a} = a^{3}.

Dividing cancels matching factors: in a5a2\tfrac{a^5}{a^2}, two of the five aa's on top cancel the two on the bottom, leaving three. When the bottom has more, the leftovers stay on the bottom, a2a6=1a4\tfrac{a^2}{a^6} = \tfrac{1}{a^4}, and the rule's answer a−4a^{-4} is the same thing, which is exactly why negative exponents mean reciprocals.

Example
Dividing like bases
x9x4=x9−4=x5,3735=32=9,12b84b3=124 b8−3=3b5.\begin{aligned} \frac{x^{9}}{x^{4}} &= x^{9-4} = x^{5}, \\[4pt] \frac{3^{7}}{3^{5}} &= 3^{2} = 9, \\[4pt] \frac{12b^{8}}{4b^{3}} &= \frac{12}{4}\,b^{8-3} = 3b^{5}. \end{aligned}
Side note
Where it goes wrong

Subtracting in the wrong order: x3x7=x4\tfrac{x^3}{x^7} = x^4. The rule is top minus bottom: x3−7=x−4=1x4x^{3 - 7} = x^{-4} = \tfrac{1}{x^4}. More factors on the bottom means the result is a fraction.

Tip

Top minus bottom — order matters. a2a6=a2−6=a−4=1a4\dfrac{a^{2}}{a^{6}} = a^{2-6} = a^{-4} = \dfrac{1}{a^{4}}. A negative result just means the base “lives” in the denominator.

Try it
Try it: quotient rule

Simplify 20m8n25m3n6\dfrac{20m^8n^2}{5m^3n^6} with positive exponents.

Show the answer

Coefficients: 205=4\tfrac{20}{5} = 4. m8−3=m5m^{8 - 3} = m^5. n2−6=n−4=1n4n^{2 - 6} = n^{-4} = \tfrac{1}{n^4}. Result: 4m5n4\dfrac{4m^5}{n^4}.

Power of a Power

Concept
Multiply the Exponents

When a power is raised to another power, keep the base and multiply the exponents.

(am)n=amn.\left(a^{m}\right)^{n} = a^{mn}.

Because (a2)3=a2⋅a2⋅a2=a6\left(a^{2}\right)^{3} = a^{2}\cdot a^{2}\cdot a^{2} = a^{6}.

(a2)3(a^2)^3 means three copies of a2a^2 multiplied together, a2⋅a2⋅a2a^2 \cdot a^2 \cdot a^2, and by the product rule that is a2+2+2=a2⋅3a^{2 + 2 + 2} = a^{2 \cdot 3}. Repeated addition of the same exponent is multiplication, which is where this rule comes from.

Example
Raising a power to a power
(x3)4=x3⋅4=x12,(22)3=26=64,(y5)2=y10.\begin{aligned} \left(x^{3}\right)^{4} &= x^{3\cdot 4} = x^{12}, \\[2pt] \left(2^{2}\right)^{3} &= 2^{6} = 64, \\[2pt] \left(y^{5}\right)^{2} &= y^{10}. \end{aligned}
Side note
Where it goes wrong

Adding instead: (x4)3=x7(x^4)^3 = x^7. Three copies of x4x^4 make x4+4+4=x12x^{4 + 4 + 4} = x^{12}.

Tip

Do not confuse the rules: (x3)4=x12\left(x^{3}\right)^{4} = x^{12} (multiply), but x3⋅x4=x7x^{3}\cdot x^{4} = x^{7} (add). Ask yourself whether the powers are being multiplied together or stacked.

Try it
Try it: stacked powers

Simplify (y3)4⋅y2(y^3)^4 \cdot y^2 and ((22)3)2\left((2^2)^3\right)^2.

Show the answer

(y3)4=y12(y^3)^4 = y^{12}, then y12⋅y2=y14y^{12} \cdot y^2 = y^{14}. ((22)3)2=22⋅3⋅2=212=4096\left((2^2)^3\right)^2 = 2^{2 \cdot 3 \cdot 2} = 2^{12} = 4096.

Power of a Product and Power of a Quotient

Concept
Distribute the Exponent

An exponent on a product or a quotient applies to every factor inside.

(ab)n=anbn,(ab)n=anbn.\left(ab\right)^{n} = a^{n}b^{n}, \qquad \left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}.

(ab)3=(ab)(ab)(ab)(ab)^3 = (ab)(ab)(ab), and multiplication can be rearranged freely, so it equals (aaa)(bbb)=a3b3(aaa)(bbb) = a^3b^3. The exponent reaches every factor, including numbers: (2x)3=23x3=8x3(2x)^3 = 2^3x^3 = 8x^3. The same rearranging does not work for sums, which is why (a+b)2≠a2+b2(a + b)^2 \ne a^2 + b^2.

Example
Distributing over products and quotients
(2x)3=23x3=8x3,(3a2b)2=32 a2⋅2 b2=9a4b2,(x4)2=x242=x216.\begin{aligned} \left(2x\right)^{3} &= 2^{3}x^{3} = 8x^{3}, \\[4pt] \left(3a^{2}b\right)^{2} &= 3^{2}\,a^{2\cdot 2}\,b^{2} = 9a^{4}b^{2}, \\[4pt] \left(\frac{x}{4}\right)^{2} &= \frac{x^{2}}{4^{2}} = \frac{x^{2}}{16}. \end{aligned}

Notice how power of a product combines with power of a power: each inside exponent gets multiplied by the outside one.

Side note
Where it goes wrong

Distributing an exponent over a sum: (x+3)2=x2+9(x + 3)^2 = x^2 + 9. The exponent spreads over products only. (x+3)2=(x+3)(x+3)=x2+6x+9(x + 3)^2 = (x + 3)(x + 3) = x^2 + 6x + 9; test x=1x = 1: 1616 versus 1010.

Tip

The coefficient gets the exponent too! A common slip is (2x)3=2x3\left(2x\right)^{3} = 2x^{3}. The correct result is 8x38x^{3}, because 23=82^{3} = 8.

Try it
Try it: distribute the power

Simplify (−3a2b)3(-3a^2b)^3 and (2xy3)4\left(\dfrac{2x}{y^3}\right)^4.

Show the answer

(−3)3=−27(-3)^3 = -27, (a2)3=a6(a^2)^3 = a^6, b3b^3: result −27a6b3-27a^6b^3. For the quotient: 24x4y12=16x4y12\tfrac{2^4x^4}{y^{12}} = \tfrac{16x^4}{y^{12}}.

Zero and Negative Exponents

Concept
The Zero Exponent

Any nonzero base raised to the 00 power equals 11:

a0=1(a≠0).a^{0} = 1 \quad (a \ne 0).

This follows from the quotient rule: a3a3=a3−3=a0\dfrac{a^{3}}{a^{3}} = a^{3-3} = a^{0}, and any number divided by itself is 11.

Walk down the powers of 22: 4,2,1,…4, 2, 1, \ldots Each step divides by 22. Continuing the same pattern past 212^1 forces 20=12^0 = 1, then 2−1=122^{-1} = \tfrac12 and 2−2=142^{-2} = \tfrac14. These definitions are the only ones that keep every exponent rule true, which is why they are chosen.

Concept
Negative Exponents

A negative exponent means take the reciprocal (flip it), then use the positive exponent:

a−n=1an,1a−n=an.a^{-n} = \frac{1}{a^{n}}, \qquad \frac{1}{a^{-n}} = a^{n}.

A factor moves between numerator and denominator by flipping the sign of its exponent.

Example
Working with zero and negative exponents
42²22¹12⁰1/22⁻¹1/42⁻²
Each step down divides by 22, straight through 20=12^0 = 1.
70=1,(5x)0=1,2−3=123=18,x−4=1x4,13−2=32=9,2x−3=2x3.\begin{aligned} 7^{0} &= 1, & (5x)^{0} &= 1, \\[2pt] 2^{-3} &= \frac{1}{2^{3}} = \frac{1}{8}, & x^{-4} &= \frac{1}{x^{4}}, \\[4pt] \frac{1}{3^{-2}} &= 3^{2} = 9, & \frac{2}{x^{-3}} &= 2x^{3}. \end{aligned}
Side note
Where it goes wrong

Moving the coefficient along with a negative exponent: 3x−2=13x23x^{-2} = \tfrac{1}{3x^2}. The exponent belongs only to xx: 3x−2=3x23x^{-2} = \tfrac{3}{x^2}. With x=1x = 1: 33, not 13\tfrac13.

Tip

A negative exponent does not make the answer negative. 2−3=182^{-3} = \tfrac{1}{8}, a positive number, not −8-8. The exponent's sign controls where the factor goes (top or bottom), not the sign of the value. And remember a0=1a^{0} = 1, never 00.

Try it
Try it: zero and negative

Write with positive exponents: 5−25^{-2}, 1x−3\dfrac{1}{x^{-3}}, a−1b−2\dfrac{a^{-1}}{b^{-2}}, (4y)0(4y)^0.

Show the answer

125\tfrac{1}{25}. x3x^3 (a negative exponent in the denominator moves up). b2a\tfrac{b^2}{a}. (4y)0=1(4y)^0 = 1 for y≠0y \ne 0.

Simplifying Combined Expressions

Concept
A Game Plan

Real problems mix several rules. A reliable order:

  • Apply power-of-a-power / power-of-a-product to clear outer exponents.
  • Multiply (add exponents) and divide (subtract exponents) like bases.
  • Rewrite any negative exponents as positive by flipping their factors.
  • Simplify the number coefficients.

A fully simplified answer usually has no negative exponents and each base appearing once.

The game plan follows the order of operations: exponents outside parentheses act first, so clear them with the power rules; then multiply and divide like bases. Doing it in that order means each rule is applied to a simple product, where the counting argument behind it plainly holds.

Example
Combining several rules

Simplify   (2x2y)34x4y\;\dfrac{\left(2x^{2}y\right)^{3}}{4x^{4}y}.

(2x2y)34x4y=23x6y34x4ypower of a product=8x6y34x4y23=8=84 x6−4 y3−1quotient rule=2x2y2.\begin{aligned} \frac{\left(2x^{2}y\right)^{3}}{4x^{4}y} &= \frac{2^{3}x^{6}y^{3}}{4x^{4}y} && \text{power of a product} \\[4pt] &= \frac{8x^{6}y^{3}}{4x^{4}y} && 2^{3}=8 \\[4pt] &= \frac{8}{4}\,x^{6-4}\,y^{3-1} && \text{quotient rule} \\[4pt] &= 2x^{2}y^{2}. \end{aligned}
Side note
Where it goes wrong

Applying the outer exponent to only one factor: (2x2y)3=2x6y3(2x^2y)^3 = 2x^6y^3. The 22 is a factor inside too: 8x6y38x^6y^3.

Example
Clearing negative exponents

Simplify   6a−2b33a2b−1\;\dfrac{6a^{-2}b^{3}}{3a^{2}b^{-1}} with only positive exponents.

6a−2b33a2b−1=63 a−2−2 b3−(−1)quotient rule=2a−4b4=2b4a4.move a−4 down\begin{aligned} \frac{6a^{-2}b^{3}}{3a^{2}b^{-1}} &= \frac{6}{3}\,a^{-2-2}\,b^{3-(-1)} && \text{quotient rule} \\[4pt] &= 2a^{-4}b^{4} \\[4pt] &= \frac{2b^{4}}{a^{4}}. && \text{move } a^{-4}\text{ down} \end{aligned}
Tip

When subtracting a negative exponent, be careful: b3−(−1)=b3+1=b4b^{3-(-1)} = b^{3+1} = b^{4}. Two minus signs make a plus.

Try it
Try it: combine

Simplify (3a−1b2)29a−4b\dfrac{(3a^{-1}b^2)^2}{9a^{-4}b} with positive exponents.

Show the answer

Numerator: 9a−2b49a^{-2}b^4. Divide: 99=1\tfrac99 = 1, a−2−(−4)=a2a^{-2 - (-4)} = a^2, b4−1=b3b^{4 - 1} = b^3. Result: a2b3a^2b^3.

Scientific Notation

Concept
The Form

Scientific notation writes a number as

c×10n,where 1≤c<10 and n is an integer.c \times 10^{n}, \qquad \text{where } 1 \le c < 10 \text{ and } n \text{ is an integer.}

The factor cc (the coefficient) has exactly one nonzero digit before the decimal point. The power of 1010 records how far the decimal point moved.

  • Large numbers use a positive exponent: 4 300=4.3×1034\,300 = 4.3\times 10^{3}.
  • Small numbers (between 00 and 11) use a negative exponent: 0.0057=5.7×10−30.0057 = 5.7\times 10^{-3}.

Scientific notation separates a number's size (the power of ten) from its digits (the coefficient). That separation is what makes multiplication easy: coefficients multiply, powers of ten add by the product rule, and a final shift restores 1≤a<101 \le a < 10.

Example
Converting to and from standard form

To scientific notation — move the decimal so one nonzero digit stays in front, and count the moves:

52 000=5.2×104moved left 4 places0.00081=8.1×10−4moved right 4 places\begin{aligned} 52\,000 &= 5.2\times 10^{4} && \text{moved left 4 places} \\ 0.00081 &= 8.1\times 10^{-4} && \text{moved right 4 places} \end{aligned}

To standard form — move the decimal the number of places the exponent says (right if positive, left if negative):

3.6×105=360 000,9.2×10−3=0.0092.\begin{aligned} 3.6\times 10^{5} &= 360\,000, \\ 9.2\times 10^{-3} &= 0.0092. \end{aligned}
Side note
Where it goes wrong

Adding exponents when adding numbers: (3×104)+(2×104)=5×108(3 \times 10^4) + (2 \times 10^4) = 5 \times 10^8. Addition keeps the power: 30,000+20,000=5×10430{,}000 + 20{,}000 = 5 \times 10^4. Only multiplication adds exponents.

Concept
Multiplying and Dividing in Scientific Notation

Handle the coefficients and the powers of 1010 separately, using the product and quotient rules on the 1010's.

(a×10m)(b×10n)=(a⋅b)×10m+n,\left(a\times 10^{m}\right)\left(b\times 10^{n}\right) = (a\cdot b)\times 10^{m+n},
a×10mb×10n=ab×10m−n.\frac{a\times 10^{m}}{b\times 10^{n}} = \frac{a}{b}\times 10^{m-n}.

At the end, adjust so the coefficient again satisfies 1≤c<101 \le c < 10.

Example
Multiplying in scientific notation

Compute (3×104)(2×105)\left(3\times 10^{4}\right)\left(2\times 10^{5}\right).

(3×104)(2×105)=(3⋅2)×104+5=6×109.\begin{aligned} \left(3\times 10^{4}\right)\left(2\times 10^{5}\right) &= (3\cdot 2)\times 10^{4+5} \\ &= 6\times 10^{9}. \end{aligned}
Example
Dividing, then fixing the coefficient

Compute 8.4×1074×103\dfrac{8.4\times 10^{7}}{4\times 10^{3}} and 1.5×1035×106\dfrac{1.5\times 10^{3}}{5\times 10^{6}}.

8.4×1074×103=8.44×107−3=2.1×104.1.5×1035×106=0.3×10−3=3×10−4.\begin{aligned} \frac{8.4\times 10^{7}}{4\times 10^{3}} &= \frac{8.4}{4}\times 10^{7-3} = 2.1\times 10^{4}. \\[6pt] \frac{1.5\times 10^{3}}{5\times 10^{6}} &= 0.3\times 10^{-3} = 3\times 10^{-4}. \end{aligned}

In the second one, 0.30.3 is too small, so we rewrite 0.3=3×10−10.3 = 3\times 10^{-1}, which drops the power of 1010 by one.

Tip

If your coefficient ends up 1010 or larger, shift the decimal left and add 11 to the exponent. If it ends up less than 11, shift right and subtract 11. Example: 12×105=1.2×10612\times 10^{5} = 1.2\times 10^{6}.

Try it
Try it: compute

Compute (6×10−3)(5×108)(6 \times 10^{-3})(5 \times 10^{8}) and 3.6×1059×10−2\dfrac{3.6 \times 10^{5}}{9 \times 10^{-2}} in scientific notation.

Show the answer

30×105=3×10630 \times 10^{5} = 3 \times 10^{6}. 3.69×105−(−2)=0.4×107=4×106\tfrac{3.6}{9} \times 10^{5 - (-2)} = 0.4 \times 10^{7} = 4 \times 10^{6}.

Exponential Growth vs. Decay (Intro)

Concept
The Pattern y=a⋅bxy = a\cdot b^{x}

Many real situations follow

y=a⋅bx,y = a\cdot b^{x},

where aa is the starting amount (the value when x=0x=0) and bb is the growth factor. The value of bb decides the behavior:

  • Growth: b>1b > 1 — the quantity gets larger as xx increases (doubling, tripling, interest).
  • Decay: 0<b<10 < b < 1 — the quantity gets smaller as xx increases (halving, cooling, fading).

Because b0=1b^{0}=1, the starting value is always y=a⋅b0=ay = a\cdot b^{0} = a.

A linear model adds the same amount each step; an exponential model multiplies by the same factor each step. Multiplying by 22 adds more each time the amount grows (5,10,20,40,805, 10, 20, 40, 80: steps of 5,10,20,405, 10, 20, 40), so doubling eventually overtakes any steady addition, as the dashed line y=5+15xy = 5 + 15x shows. A factor between 00 and 11 takes away a fixed fraction each step, so the decrease shrinks and the amount approaches zero without reaching it.

Example
Evaluating a growth model

A colony starts at 55 bacteria and doubles each hour: y=5⋅2xy = 5\cdot 2^{x}, with xx in hours. Here b=2>1b = 2 > 1, so it is growth.

x=0:y=5⋅20=5⋅1=5,x=1:y=5⋅21=10,x=3:y=5⋅23=5⋅8=40.\begin{aligned} x=0:\quad y &= 5\cdot 2^{0} = 5\cdot 1 = 5, \\ x=1:\quad y &= 5\cdot 2^{1} = 10, \\ x=3:\quad y &= 5\cdot 2^{3} = 5\cdot 8 = 40. \end{aligned}
Side note
Where it goes wrong

Reading y=80(0.5)xy = 80(0.5)^x as losing 0.50.5 each bounce. The factor multiplies: 80,40,20,10,…80, 40, 20, 10, \ldots Each bounce loses half of the current height, not a fixed amount.

Example
Evaluating a decay model
xy123451020304050607080y = 5·2ˣy = 80(1/2)ˣ
Doubling from 55 (rising), halving from 8080 (falling), and the linear y=5+15xy = 5 + 15x (dashed), which doubling overtakes between x=3x = 3 and x=4x = 4.

A ball bounces to 12\tfrac{1}{2} its previous height, starting at 8080 cm: y=80⋅(12)xy = 80\cdot\left(\tfrac{1}{2}\right)^{x}. Here b=12b = \tfrac{1}{2}, and 0<b<10 < b < 1, so it is decay.

x=0:y=80⋅(12)0=80,x=1:y=80⋅12=40,x=2:y=80⋅14=20.\begin{aligned} x=0:\quad y &= 80\cdot\left(\tfrac{1}{2}\right)^{0} = 80, \\ x=1:\quad y &= 80\cdot\tfrac{1}{2} = 40, \\ x=2:\quad y &= 80\cdot\tfrac{1}{4} = 20. \end{aligned}
Tip

Read bb to tell the story: b>1b>1 means the amount multiplies up (growth); 0<b<10<b<1 means it shrinks toward zero (decay). Either way, plug in xx and use your exponent skills to evaluate.

Try it
Try it: model it

A phone worth $800\$800 loses 25%25\% of its value each year. Write the model and find its value after 22 years. Is it growth or decay?

Show the answer

Keeping 75%75\% each year: y=800(0.75)xy = 800(0.75)^x. After 22 years: 800⋅0.5625=450800 \cdot 0.5625 = 450 dollars. The factor 0.750.75 is between 00 and 11, so it is decay.

Going Deeper: Advanced Exponents

Concept
Solving Exponential Equations by Matching Bases

If two powers with the same base are equal, then their exponents must be equal:

bm=bn⟹m=n(b>0, b≠1).b^{m} = b^{n} \quad\Longrightarrow\quad m = n \qquad (b>0,\ b\ne 1).

To solve an equation like b □=Nb^{\,\square} = N, rewrite both sides as powers of one common base, then set the exponents equal and solve that (usually linear) equation.

  • Ask: “What base makes both sides simple?” Often it is the smaller of the two numbers.
  • Use the power-of-a-power rule to fold the outside exponent in: (bk)x=bkx\left(b^{k}\right)^{x} = b^{kx}.
Example
Matching bases to solve

Solve each equation for xx.

2x=32⇒2x=25⇒x=5.9x=27⇒(32)x=33⇒32x=33,  2x=3,  x=32.5 x+1=125⇒5 x+1=53⇒x+1=3,  x=2.\begin{aligned} 2^{x} &= 32 &&\Rightarrow&& 2^{x} = 2^{5} &&\Rightarrow&& x = 5. \\[4pt] 9^{x} &= 27 &&\Rightarrow&& \left(3^{2}\right)^{x} = 3^{3} &&\Rightarrow&& 3^{2x}=3^{3},\ \ 2x=3,\ \ x=\tfrac{3}{2}. \\[4pt] 5^{\,x+1} &= 125 &&\Rightarrow&& 5^{\,x+1} = 5^{3} &&\Rightarrow&& x+1 = 3,\ \ x = 2. \end{aligned}

Each time: write both sides with the same base, then just compare exponents.

Concept
Rational Exponents: the Bridge to Radicals

A fractional exponent is a compact way to write a root. For a≥0a \ge 0 and positive integers m,nm,n:

a1/n=an,am/n=amn=(an)m.a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}.

Read the fraction as “root on the bottom, power on the top”: the denominator nn is the index of the radical, and the numerator mm is the ordinary power. You may take the root first or the power first — the answer is the same, but taking the root first keeps the numbers small.

Example
Rational exponents both ways

Evaluate, taking the root first.

82/3=(83)2=22=4,163/4=(164)3=23=8,27−1/3=1271/3=1273=13.\begin{aligned} 8^{2/3} &= \left(\sqrt[3]{8}\right)^{2} = 2^{2} = 4, \\[4pt] 16^{3/4} &= \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8, \\[4pt] 27^{-1/3} &= \frac{1}{27^{1/3}} = \frac{1}{\sqrt[3]{27}} = \frac{1}{3}. \end{aligned}

The last one combines two ideas: the negative sign flips it, and the 1/31/3 takes a cube root.

Tip

All the ordinary exponent rules still work with fractional exponents. For instance a⋅a3=a1/2⋅a1/3=a1/2+1/3=a5/6\sqrt{a}\cdot\sqrt[3]{a} = a^{1/2}\cdot a^{1/3} = a^{1/2+1/3} = a^{5/6}. Rewriting radicals as powers is often the fastest way to combine them.

Concept
Comparing Powers of Different Bases

You cannot just eyeball which of 2402^{40} or 3303^{30} is larger. The trick is to make either the exponents match or the bases match, then compare.

  • Match the exponents: factor the exponents to pull out a common one, then compare the resulting bases.
  • Match the bases: rewrite both as powers of a shared base when possible.

Once one part agrees, the larger base (or larger exponent) wins.

Example
Which power is bigger?

Compare 2402^{40} and 3303^{30} by giving them a common exponent of 1010.

240=(24)10=1610,330=(33)10=2710.\begin{aligned} 2^{40} &= \left(2^{4}\right)^{10} = 16^{10}, \\[4pt] 3^{30} &= \left(3^{3}\right)^{10} = 27^{10}. \end{aligned}

Both are raised to the 1010th power, and 27>1627 > 16, so 2710>161027^{10} > 16^{10}. Therefore 330>2403^{30} > 2^{40}.

Concept
Units-Digit Cycles

The units digit (last digit) of a power repeats in a short cycle as the exponent grows, because only the previous units digit affects the next one. For base 77:

71 ⁣= ⁣7,72 ⁣= ⁣49,73 ⁣= ⁣343,74 ⁣= ⁣2401,75 ⁣= ⁣16807, …7^{1}\!=\!7,\quad 7^{2}\!=\!49,\quad 7^{3}\!=\!343,\quad 7^{4}\!=\!2401,\quad 7^{5}\!=\!16807,\ \dots

so the units digits run 7, 9, 3, 1, 7, 9, 3, 1,…7,\,9,\,3,\,1,\,7,\,9,\,3,\,1,\dots — a cycle of length 44. To find the units digit of 7k7^{k}, divide kk by the cycle length and use the remainder to pick the spot in the cycle (a remainder of 00 lands on the last entry).

Example
Finding a units digit

Find the units digit of 71007^{100}.

Cycle of units digits: 7, 9, 3, 1(length 4).100÷4=25 remainder 0 ⇒ use the 4th (last) entry.Units digit of 7100=1.\begin{aligned} &\text{Cycle of units digits: } 7,\,9,\,3,\,1 \quad(\text{length }4). \\[2pt] &100 \div 4 = 25 \text{ remainder } 0 \ \Rightarrow\ \text{use the 4th (last) entry.} \\[2pt] &\text{Units digit of } 7^{100} = 1. \end{aligned}

Check the idea on a small case: 100≡0(mod4)100 \equiv 0 \pmod 4 matches 74=24017^{4}=2401, which indeed ends in 11.

Concept
Growth and Decay by a Percent Rate

The model y=a⋅bxy = a\cdot b^{x} often comes from a percent change per step. Writing the growth factor as b=1±rb = 1 \pm r (with rr a decimal rate) separates the two cases:

Growth: y=a (1+r)x,Decay: y=a (1−r)x.\text{Growth: } y = a\,(1+r)^{x}, \qquad \text{Decay: } y = a\,(1-r)^{x}.

Here aa is the starting amount and rr is the rate per period. Growth uses 1+r>11+r>1; decay uses 1−r1-r between 00 and 11.

Example
A percent-growth model

A $2000 investment grows 5%5\% per year, so r=0.05r = 0.05 and b=1.05b = 1.05:

y=2000 (1.05)x.y = 2000\,(1.05)^{x}.
x=0:y=2000 (1.05)0=2000,x=1:y=2000 (1.05)1=2100,x=2:y=2000 (1.05)2=2000 (1.1025)=2205.\begin{aligned} x=0:\quad y &= 2000\,(1.05)^{0} = 2000, \\ x=1:\quad y &= 2000\,(1.05)^{1} = 2100, \\ x=2:\quad y &= 2000\,(1.05)^{2} = 2000\,(1.1025) = 2205. \end{aligned}

A 12%12\% decay would instead use b=1−0.12=0.88b = 1 - 0.12 = 0.88.

Concept
The Logarithm: Undoing an Exponent (Preview)

Exponentiation asks “what do I get when I raise bb to the power xx?” The logarithm asks the reverse: “what power of bb gives me yy?” They are inverse operations:

bx=y⟺log⁡by=x(b>0, b≠1).b^{x} = y \quad\Longleftrightarrow\quad \log_{b} y = x \qquad (b>0,\ b\ne 1).

So log⁡by\log_{b} y is simply the exponent you would put on bb to land on yy. For example, log⁡28=3\log_{2} 8 = 3 because 23=82^{3}=8, and log⁡101000=3\log_{10} 1000 = 3 because 103=100010^{3}=1000. This is the tool that solves 2x=102^{x}=10, where the bases cannot be matched by hand.

Concept
Adding and Subtracting in Scientific Notation

For multiplying and dividing you handle the coefficients and the powers of 1010 separately. But to add or subtract, the powers of 1010 must first match. Rewrite the smaller-exponent number so both share the larger exponent, then add or subtract the coefficients:

a×10n±b×10n=(a±b)×10n.a\times 10^{n} \pm b\times 10^{n} = (a \pm b)\times 10^{n}.

Finally, adjust the coefficient back into the range 1≤c<101 \le c < 10 if needed.

Example
Adding with a common power of ten

Compute (3×104)+(2×103)\left(3\times 10^{4}\right) + \left(2\times 10^{3}\right).

3×104+2×103=3×104+0.2×104match the exponents=(3+0.2)×104add the coefficients=3.2×104.\begin{aligned} 3\times 10^{4} + 2\times 10^{3} &= 3\times 10^{4} + 0.2\times 10^{4} && \text{match the exponents} \\[4pt] &= (3 + 0.2)\times 10^{4} && \text{add the coefficients} \\[4pt] &= 3.2\times 10^{4}. \end{aligned}

Rewriting 2×1032\times 10^{3} as 0.2×1040.2\times 10^{4} lets both terms share 10410^{4} before adding.

Tip

Never add the powers of 1010 when you are adding numbers — that is the rule for multiplying. For a sum, the exponents must already be equal; only the coefficients get added.

Why It Works

Follow the pattern 23=82^3 = 8, 22=42^2 = 4, 21=22^1 = 2: each step down divides by 22. Continuing, 20=12^0 = 1, 2−1=122^{-1} = \tfrac12, 2−2=142^{-2} = \tfrac14. The definitions of zero and negative exponents are the only ones that keep the rule am/an=am−na^m/a^n = a^{m-n} true everywhere. Exponential growth beats linear growth because its increments themselves grow: doubling adds more each step, while a line adds the same amount forever.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
The exponent rules
aman=am+n,aman=am−n,(am)n=amn,a0=1,a−n=1ana^m a^n = a^{m+n},\qquad \frac{a^m}{a^n}=a^{m-n},\qquad (a^m)^n = a^{mn},\qquad a^0 = 1,\qquad a^{-n}=\frac{1}{a^{n}}

What it means. Multiplying powers adds exponents; dividing subtracts them; a power of a power multiplies them.

Example. 23⋅24=27=1282^3\cdot 2^4=2^7=128, and 2522=23=8\dfrac{2^5}{2^2}=2^3=8.

Why it works. amana^m a^n writes aa down mm times then nn more times — m+nm+n copies in all. Division cancels copies, leaving m−nm-n. For a0a^0: amam=am−m=a0\tfrac{a^m}{a^m}=a^{m-m}=a^0, and any nonzero number over itself is 11. For negatives: a−n=a0−n=a0an=1ana^{-n}=a^{0-n}=\tfrac{a^0}{a^n}=\tfrac{1}{a^n}.

Tip. The rules only combine powers of the same base. 23⋅322^3\cdot 3^2 does not simplify by adding exponents.