Exponents & Roots

In $7^{3}$, the base is $\mathbf{7}$ and the exponent is $\mathbf{3}$. It means $7 \times 7 \times 7$. Evaluating (working it out): $7 \times 7 = 49$, then $49 \times 7 = \mathbf{343}$.

Write $4 \times 4 \times 4 \times 4 \times 4$ as a power. The base is $4$ and it appears $5$ times, so this is $\mathbf{4^{5}}$. (Its value is $1024$, but writing $4^{5}$ is much shorter.)

$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = \mathbf{32}$.   $3^{4} = 3 \times 3 \times 3 \times 3 = \mathbf{81}$.   $6^{2} = 6 \times 6 = \mathbf{36}$.

$10^{1} = 10$,   $10^{2} = 100$,   $10^{3} = 1000$,   $10^{6} = 1{,}000{,}000$ (one million). Count the zeros: the exponent tells you exactly how many.

$2^{3} \cdot 2^{4} = 2^{3+4} = 2^{7} = \mathbf{128}$. Check the long way: $2^{3} = 8$ and $2^{4} = 16$, and $8 \times 16 = 128$. It matches.

Quotient: $\dfrac{5^{6}}{5^{2}} = 5^{6-2} = 5^{4} = \mathbf{625}$.   Power of a power: $(3^{2})^{3} = 3^{2 \times 3} = 3^{6} = \mathbf{729}$.

$7^{0} = 1$,   $100^{0} = 1$,   $2^{0} = 1$. It does not matter how big the base is --- as long as it is not zero, a zero exponent gives $\mathbf{1}$.

Watch the powers of $2$ shrink by half as the exponent drops: $2^{3}=8$, $2^{2}=4$, $2^{1}=2$, and the next step is $2^{0}=\mathbf{1}$. The pattern lands exactly on $1$.

Write $52{,}000$ in scientific notation. Put the decimal after the first nonzero digit: $5.2$. To get from $5.2$ back to $52{,}000$ the decimal moves $4$ places to the right, so the exponent is $4$: $52{,}000 = \mathbf{5.2 \times 10^{4}}$.

Write $0.00047$ in scientific notation. The first nonzero digit gives $4.7$; the decimal must move $4$ places to the left to make the number small, so the exponent is negative: $0.00047 = \mathbf{4.7 \times 10^{-4}}$. Converting back: $3 \times 10^{5}$ means move the decimal $5$ places right: $\mathbf{300{,}000}$.

$\sqrt{49} = 7$ because $7^{2} = 49$.   $\sqrt{81} = 9$ because $9^{2} = 81$.   $\sqrt{144} = 12$ because $12^{2} = 144$.

$1^{2}=1$, $2^{2}=4$, $3^{2}=9$, $4^{2}=16$, $5^{2}=25$, $6^{2}=36$, $7^{2}=49$, $8^{2}=64$, $9^{2}=81$, $10^{2}=100$, $11^{2}=121$, $12^{2}=144$. Knowing these by heart makes square roots quick.

Estimate $\sqrt{50}$. The nearest perfect squares are $49$ and $64$, so $\sqrt{49} = 7$ and $\sqrt{64} = 8$. Since $50$ is between $49$ and $64$, we know $7 < \sqrt{50} < 8$. Because $50$ is very close to $49$, $\sqrt{50}$ is just a bit more than $\mathbf{7}$.

Estimate $\sqrt{30}$. It sits between $\sqrt{25}=5$ and $\sqrt{36}=6$, so $5 < \sqrt{30} < 6$. Since $30$ is closer to $25$ than to $36$, $\sqrt{30}$ is closer to $\mathbf{5}$ (about $5.5$).

Evaluate $4 + 3 \times 2^{3}$. Exponent first: $2^{3} = 8$, giving $4 + 3 \times 8$. Multiply: $3 \times 8 = 24$. Add: $4 + 24 = \mathbf{28}$.

Evaluate $\sqrt{9 + 16} + 5^{2}$. Under the root first: $9 + 16 = 25$, and $\sqrt{25} = 5$. Then $5^{2} = 25$. Add: $5 + 25 = \mathbf{30}$. (Note $\sqrt{9+16} \neq \sqrt{9}+\sqrt{16}$, which would give $3+4=7$.)

A square garden has sides of $9$ meters. Its area is $9^{2} = 9 \times 9 = \mathbf{81}$ square meters. Working backward: if a square tile has an area of $64$ square centimeters, each side is $\sqrt{64} = \mathbf{8}$ centimeters.

A kilometer is $10^{3} = 1000$ meters. So $5$ kilometers is $5 \times 1000 = \mathbf{5000}$ meters. A single bacterium about $2 \times 10^{-6}$ meters wide is tiny --- the negative exponent tells us it is far smaller than one meter.

Evaluate $2^{-3}$ and $10^{-2}$. Flip each into “one over the positive power”:

Check with the shrinking pattern: $2^{2}=4$, $2^{1}=2$, $2^{0}=1$, $2^{-1}=\tfrac{1}{2}$, $2^{-2}=\tfrac{1}{4}$, $2^{-3}=\mathbf{\tfrac{1}{8}}$. Each step still divides by the base, right on through zero into the negatives.

Compare $4^{5}$ and $5^{4}$. Evaluate each:

So $4^{5} > 5^{4}$, even though $5$ is the bigger base --- the larger exponent did more work here. Another famous near-tie: $2^{10} = 1024$ while $10^{3} = 1000$, so $2^{10}$ is just barely larger. This is why programmers say a “kilobyte” ($2^{10}$ bytes) is about a thousand.

The ones digit of a power repeats in a short cycle, so you can find it without computing the whole number. Watch the units digits of the powers of $2$:

The last digits run $2, 4, 8, 6,\; 2, 4, 8, 6,\ldots$ --- a cycle of length $4$. To find the units digit of $2^{30}$, divide the exponent by $4$: $30 = 4\times 7 + 2$, remainder $2$. The $2$nd entry in the cycle is $\mathbf{4}$, so $2^{30}$ ends in $4$. Powers of $3$ cycle $3, 9, 7, 1$; powers of $7$ cycle $7, 9, 3, 1$.

Product: $(3\times 10^{4})\times(2\times 10^{3})$. Multiply fronts $3\times 2 = 6$; add exponents $4+3 = 7$:

Product needing a fix-up: $(4\times 10^{5})(3\times 10^{-2}) = 12\times 10^{3}$. Since $12$ is too big, rewrite $12 = 1.2\times 10^{1}$: the answer is $1.2\times 10^{1}\times 10^{3} = \mathbf{1.2\times 10^{4}}$. Sum: $3\times 10^{4} + 2\times 10^{3}$. Match the powers: $2\times 10^{3} = 0.2\times 10^{4}$, so $3\times 10^{4} + 0.2\times 10^{4} = \mathbf{3.2\times 10^{4}}$.

We already know $7 < \sqrt{50} < 8$, and $50$ is just above $49$, so try $7.1$: $7.1^{2} = 50.41$, a little too big. Try $7.07$: $7.07^{2} = 49.98$, just under $50$. So $\sqrt{50} \approx \mathbf{7.07}$. The same “guess, square, adjust” idea works for any root: to estimate $\sqrt{20}$, note $4.4^{2}=19.36$ (low) and $4.5^{2}=20.25$ (high), so $\sqrt{20}\approx 4.47$.