Exponential & Logarithmic Functions

Study Sheet

Exponential & Logarithmic Functions

Everything for Topic 8 on one desk reference

Exponential Functions y=abxy = ab^x

Concept
Anatomy of an Exponential

An exponential function has the form

y=abx,a≠0,  b>0,  b≠1.y = ab^x, \qquad a \neq 0,\ \ b > 0,\ \ b \neq 1.

The variable lives in the exponent. Here aa is the initial value (the yy-intercept, since b0=1b^0 = 1 gives y=ay=a), and bb is the base or growth factor.

  • If b>1b > 1: exponential growth (curve rises to the right).
  • If 0<b<10 < b < 1: exponential decay (curve falls to the right).

For a>0a>0: domain is all real numbers; range is y>0y > 0; the line y=0y = 0 is a horizontal asymptote the curve approaches but never touches.

A logarithm is an exponent: log⁡ba=c\log_b a = c means bc=ab^c = a. The log laws (product to sum, quotient to difference, power to multiple) are the exponent laws read backwards, and change of base lets any calculator compute any log. Exponential growth and decay, compound interest and half-lives are all y=a⋅bty = a\cdot b^t solved with logs.

Each unit step in xx multiplies yy by the same factor bb, just as a linear function adds the same amount. That is why b>1b > 1 gives growth that keeps accelerating, 0<b<10 < b < 1 gives decay toward 00, and the curve never reaches 00: multiplying a nonzero number by bb can never produce zero.

Growth y=2xy=2^x (solid) rises to the right; decay y=(12)xy=\left(\tfrac12\right)^x (dashed) falls. Both share the asymptote y=0y=0 and pass through (0,1)(0,1).

Concept
Transformations of y=bxy = b^x

Starting from the parent y=bxy = b^x, the function y=a b x−h+ky = a\,b^{\,x-h} + k is:

  • shifted right hh and up kk;
  • stretched by ∣a∣|a| (reflected over the xx-axis if a<0a<0);
  • its horizontal asymptote moves to y=ky = k, and the range becomes y>ky > k (or y<ky<k if a<0a<0).
Example
Worked Example: Reading a Transformation

Describe y=2x−2y = 2^{x} - 2 and give its asymptote and range. It is the parent y=2xy=2^x shifted down 2. The asymptote drops to y=−2y=-2, so the range is y>−2y > -2. The yy-intercept is 20−2=−12^0 - 2 = -1.

Side note
Where it goes wrong

Writing log⁡(a+b)=log⁡a+log⁡b\log(a + b) = \log a + \log b. Logs turn PRODUCTS into sums; there is no rule for the log of a sum.

Tip

Tip: To find aa and bb from two points, use the yy-intercept for aa, then divide a second output by aa to get bxb^x. From (0,3)(0,3) and (1,12)(1,12): a=3a=3, and 12=3b112 = 3b^1 gives b=4b=4, so y=3⋅4xy = 3\cdot 4^x.

Try it
Try it: build the model

An exponential function passes through (0,5)(0, 5) and (2,45)(2, 45). Find y=abxy = ab^x and yy at x=3x = 3.

Show the answer

a=5a = 5 from the intercept. Then 45=5b245 = 5b^2, so b2=9b^2 = 9 and b=3b = 3 (the base must be positive). y=5⋅3xy = 5 \cdot 3^x, and y(3)=5⋅27=135y(3) = 5 \cdot 27 = 135.

The Number ee & Continuous Growth

Concept
Euler's Number ee

The constant e≈2.71828e \approx 2.71828 is the natural base of growth. It arises as the limit of (1+1n)n\left(1+\tfrac{1}{n}\right)^n as n→∞n \to \infty. The function y=exy = e^x is a growth curve just like any bxb^x with b>1b>1; it models continuous change.

Compounding more often pays interest on interest sooner, but the extra gain shrinks each time the period is split: (1+1n)n\left(1 + \tfrac1n\right)^n climbs 2,2.25,2.37,…2, 2.25, 2.37, \ldots and levels off at e≈2.718e \approx 2.718. That limit is what "compounded continuously" means, which is why A=PertA = Pe^{rt} is the ceiling for every compounding schedule.

Concept
Interest Formulas

For principal PP, annual rate rr (as a decimal), and time tt in years:

compounded n times/year:A=P(1+rn)nt,compounded continuously:A=Pert.\text{compounded } n \text{ times/year:}\quad A = P\left(1 + \frac{r}{n}\right)^{nt}, \qquad \text{compounded continuously:}\quad A = P e^{rt}.

Use n=1n=1 (annually), 44 (quarterly), 1212 (monthly), 365365 (daily).

Example
Worked Example: Two Kinds of Compounding
tA−112345110012001300$1349.86
$1000\$1000 at 6%6\% compounded continuously, against simple interest (dashed).

Invest $1000 at 6%6\% for 55 years.

monthly: A=1000(1+0.0612)12⋅5=1000(1.005)60≈$1348.85,continuous: A=1000 e0.06⋅5=1000 e0.3≈$1349.86.\begin{aligned} \text{monthly: } A &= 1000\left(1 + \tfrac{0.06}{12}\right)^{12\cdot 5} = 1000(1.005)^{60} \approx \$1348.85,\\ \text{continuous: } A &= 1000\,e^{0.06\cdot 5} = 1000\,e^{0.3} \approx \$1349.86. \end{aligned}

Continuous compounding earns slightly more — it is the limit as n→∞n\to\infty.

Side note
Where it goes wrong

Using the percent instead of the decimal: 1000e6⋅51000e^{6 \cdot 5}, an absurd amount. The rate is r=0.06r = 0.06: 1000e0.3≈1349.861000e^{0.3} \approx 1349.86.

Tip

Remember: more frequent compounding always yields more, but the gains shrink and level off at the continuous value PertPe^{rt}.

Try it
Try it: compare

Invest $2000\$2000 at 5%5\% for 1010 years. Find the amount compounded quarterly and continuously.

Show the answer

Quarterly: 2000(1+0.054)40=2000(1.0125)40≈$3287.242000\left(1 + \tfrac{0.05}{4}\right)^{40} = 2000(1.0125)^{40} \approx \$3287.24. Continuously: 2000e0.5≈$3297.442000e^{0.5} \approx \$3297.44.

Logarithms: Definition & Evaluating

Concept
A Logarithm is an Exponent

A logarithm answers the question “to what power?” The definition is the inverse relationship

log⁡by=x⟺bx=y,b>0, b≠1, y>0.\log_b y = x \quad\Longleftrightarrow\quad b^x = y, \qquad b>0,\ b\neq 1,\ y>0.

So log⁡by\log_b y is the exponent you put on bb to get yy. To evaluate, rewrite in exponential form.

A logarithm exists to name an exponent: 2x=322^x = 32 asks for an exponent, and the answer, 55, is written log⁡232\log_2 32. So every log fact is an exponent fact read backward: log⁡b1=0\log_b 1 = 0 because b0=1b^0 = 1, and log⁡bb=1\log_b b = 1 because b1=bb^1 = b. Logs of negatives or zero are undefined because no power of a positive base is negative or zero.

Example
Worked Example: Evaluating
log⁡232=5because 25=32,log⁡525=2because 52=25,log⁡84=23because 82/3=(23)2/3=22=4,log⁡b1=0because b0=1.\begin{aligned} \log_2 32 &= 5 &&\text{because } 2^5 = 32,\\ \log_5 25 &= 2 &&\text{because } 5^2 = 25,\\ \log_8 4 &= \tfrac{2}{3} &&\text{because } 8^{2/3} = (2^3)^{2/3} = 2^2 = 4,\\ \log_b 1 &= 0 &&\text{because } b^0 = 1. \end{aligned}
Side note
Where it goes wrong

Evaluating log⁡84\log_8 4 as 12\tfrac12 because "44 is half of 88." The log asks for the exponent: 82/3=48^{2/3} = 4, so log⁡84=23\log_8 4 = \tfrac23.

Concept
Common and Natural Logs

Two bases are so useful they get their own notation:

  • Common log: log⁡x\log x means log⁡10x\log_{10} x (base 10).
  • Natural log: ln⁡x\ln x means log⁡ex\log_e x (base ee).

Inverse facts: log⁡10k=k\log 10^k = k, ln⁡ek=k\ln e^k = k, 10log⁡x=x10^{\log x}=x, eln⁡x=xe^{\ln x}=x.

Tip

Tip: log⁡1000=3\log 1000 = 3 and ln⁡e2=2\ln e^2 = 2 take no calculator — read them straight from the definition.

Try it
Try it: evaluate

Evaluate log⁡381\log_3 81, log⁡4116\log_4 \tfrac1{16}, log⁡255\log_{25} 5 and ln⁡e7\ln e^7.

Show the answer

34=813^4 = 81, so 44. 4−2=1164^{-2} = \tfrac1{16}, so −2-2. 251/2=525^{1/2} = 5, so 12\tfrac12. ln⁡e7=7\ln e^7 = 7.

Graphs of Logarithmic Functions

Concept
The Inverse of an Exponential

Because log⁡bx\log_b x undoes bxb^x, the graph of y=log⁡bxy = \log_b x is the reflection of y=bxy = b^x across the line y=xy = x. Consequences for y=log⁡bxy=\log_b x (with b>1b>1):

  • Domain x>0x > 0; range all real numbers.
  • Vertical asymptote at x=0x = 0 (the yy-axis).
  • passes through (1,0)(1,0) since log⁡b1=0\log_b 1 = 0.

A shift y=log⁡b(x−h)y = \log_b(x-h) moves the asymptote to x=hx = h and the domain to x>hx > h.

Swapping the coordinates of every point on y=2xy = 2^x gives y=log⁡2xy = \log_2 x: (1,2)(1, 2) becomes (2,1)(2, 1), (0,1)(0, 1) becomes (1,0)(1, 0). Swapping coordinates is reflecting across y=xy = x, so everything about the log graph is the exponential's feature turned sideways: the horizontal asymptote y=0y = 0 becomes the vertical asymptote x=0x = 0.

y=log⁡2xy=\log_2 x (dashed) is the mirror image of y=2xy=2^x (solid) across y=xy=x. Its vertical asymptote is the yy-axis, x=0x=0.

Example
Worked Example: Domain of a Shifted Log

For y=log⁡2(x−3)y = \log_2(x - 3), the inside must be positive: x−3>0x - 3 > 0, so the domain is x>3x > 3 and the vertical asymptote is x=3x = 3.

Side note
Where it goes wrong

Giving y=log⁡2(x−3)y = \log_2(x - 3) the domain x>0x > 0. The inside must be positive: x−3>0x - 3 > 0, so x>3x > 3, and the asymptote moves to x=3x = 3.

Try it
Try it: shifted log

For y=log⁡3(x+2)−1y = \log_3(x + 2) - 1, give the domain, the vertical asymptote and the xx-intercept.

Show the answer

Domain: x+2>0x + 2 > 0, so x>−2x > -2; asymptote x=−2x = -2. xx-intercept: log⁡3(x+2)=1\log_3(x + 2) = 1, so x+2=3x + 2 = 3 and x=1x = 1.

Properties of Logarithms

Concept
The Three Laws (same base bb)
Product:log⁡b(MN)=log⁡bM+log⁡bN,Quotient:log⁡b ⁣(MN)=log⁡bM−log⁡bN,Power:log⁡b(Mp)=p log⁡bM.\begin{aligned} \textbf{Product:}\quad & \log_b(MN) = \log_b M + \log_b N,\\ \textbf{Quotient:}\quad & \log_b\!\left(\tfrac{M}{N}\right) = \log_b M - \log_b N,\\ \textbf{Power:}\quad & \log_b(M^p) = p\,\log_b M. \end{aligned}

These convert products into sums and exponents into multipliers. Reading them left-to-right expands; right-to-left condenses.

Each law is an exponent law in disguise. If M=bmM = b^m and N=bnN = b^n, then MN=bm+nMN = b^{m + n}, so log⁡b(MN)=m+n=log⁡bM+log⁡bN\log_b(MN) = m + n = \log_b M + \log_b N: multiplying powers adds exponents. The quotient and power laws come the same way from bmbn=bm−n\tfrac{b^m}{b^n} = b^{m - n} and (bm)p=bmp(b^m)^p = b^{mp}.

Concept
Change of Base

To evaluate a log in any base with a calculator:

log⁡bM=log⁡Mlog⁡b=ln⁡Mln⁡b.\log_b M = \frac{\log M}{\log b} = \frac{\ln M}{\ln b}.
Example
Worked Example: Expanding

Expand log⁡ ⁣x3yz2\displaystyle \log\!\frac{x^3\sqrt{y}}{z^2} completely.

log⁡x3yz2=log⁡(x3y)−log⁡z2quotient=log⁡x3+log⁡y1/2−log⁡z2product=3log⁡x+12log⁡y−2log⁡zpower\begin{aligned} \log\frac{x^3\sqrt{y}}{z^2} &= \log\big(x^3\sqrt{y}\big) - \log z^2 &&\text{quotient}\\ &= \log x^3 + \log y^{1/2} - \log z^2 &&\text{product}\\ &= 3\log x + \tfrac{1}{2}\log y - 2\log z &&\text{power} \end{aligned}
Side note
Where it goes wrong

Splitting a log of a sum: log⁡(x2+y)=2log⁡x+log⁡y\log(x^2 + y) = 2\log x + \log y. There is no law for a sum inside a log. Test with x=10x = 10, y=100y = 100: log⁡200≈2.30\log 200 \approx 2.30, but 2+2=42 + 2 = 4.

Example
Worked Example: Condensing

Condense 2ln⁡x+ln⁡y−ln⁡32\ln x + \ln y - \ln 3 into a single logarithm.

2ln⁡x+ln⁡y−ln⁡3=ln⁡x2+ln⁡y−ln⁡3power=ln⁡(x2y)−ln⁡3product=ln⁡ ⁣x2y3quotient\begin{aligned} 2\ln x + \ln y - \ln 3 &= \ln x^2 + \ln y - \ln 3 &&\text{power}\\ &= \ln(x^2 y) - \ln 3 &&\text{product}\\ &= \ln\!\frac{x^2 y}{3} &&\text{quotient} \end{aligned}
Tip

Careful: there is no rule for log⁡(M+N)\log(M+N) — you can only split products and quotients, never sums or differences inside the log. And log⁡Mlog⁡N≠log⁡MN\dfrac{\log M}{\log N} \neq \log\dfrac{M}{N}.

Try it
Try it: expand and condense

Expand log⁡28x3y\log_2\dfrac{8x^3}{\sqrt{y}}, then condense ln⁡12−2ln⁡2+ln⁡5\ln 12 - 2\ln 2 + \ln 5.

Show the answer

log⁡28+3log⁡2x−12log⁡2y=3+3log⁡2x−12log⁡2y\log_2 8 + 3\log_2 x - \tfrac12\log_2 y = 3 + 3\log_2 x - \tfrac12\log_2 y. Condense: ln⁡12⋅522=ln⁡15\ln\tfrac{12 \cdot 5}{2^2} = \ln 15.

Solving Exponential Equations

Concept
Two Strategies
  • Common base: rewrite both sides with the same base, then set the exponents equal.
  • Take a log: if a common base is awkward, isolate the power and apply log⁡\log or ln⁡\ln to both sides, then use the power rule to bring the exponent down.
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)

Taking a logarithm of both sides is allowed because equal positive numbers have equal logs, and it is useful because the power law moves the unknown exponent down to an ordinary factor: ln⁡(2x)=xln⁡2\ln(2^x) = x\ln 2. A common base does the same job without logs, since bm=bnb^m = b^n forces m=nm = n for b>0b > 0, b≠1b \ne 1.

Example
Worked Example: Common Base

Solve 4x+1=324^{x+1} = 32. Write each side as a power of 22:

(22)x+1=25 ⇒ 22x+2=25 ⇒ 2x+2=5 ⇒ x=32.(2^2)^{x+1} = 2^5 \ \Rightarrow\ 2^{2x+2} = 2^5 \ \Rightarrow\ 2x+2 = 5 \ \Rightarrow\ x = \tfrac{3}{2}.
Side note
Where it goes wrong

Taking the log before isolating the power: ln⁡(3⋅2x)=xln⁡6\ln(3 \cdot 2^x) = x\ln 6. The power rule applies only to 2x2^x; either isolate first (2x=5032^x = \tfrac{50}{3}) or split: ln⁡3+xln⁡2=ln⁡50\ln 3 + x\ln 2 = \ln 50.

Example
Worked Example: Taking a Log
xy−112341020304050x ≈ 4.06
y=3⋅2xy = 3 \cdot 2^x reaches 5050 at x=log⁡2503≈4.06x = \log_2 \tfrac{50}{3} \approx 4.06.

Solve 3⋅2x=503\cdot 2^{x} = 50. Isolate the power, then take ln⁡\ln of both sides.

2x=503ln⁡ ⁣(2x)=ln⁡ ⁣503take the natural logxln⁡2=ln⁡ ⁣503power rule brings x downx=ln⁡(50/3)ln⁡2=log⁡2 ⁣503≈4.059exact, then rounded\begin{aligned} 2^x &= \frac{50}{3}\\ \ln\!\left(2^x\right) &= \ln\!\frac{50}{3} &&\text{take the natural log}\\ x\ln 2 &= \ln\!\frac{50}{3} &&\text{power rule brings } x \text{ down}\\ x &= \frac{\ln(50/3)}{\ln 2} = \log_2\!\frac{50}{3} \approx 4.059 &&\text{exact, then rounded} \end{aligned}
Tip

Leave it exact, then round. An exact answer like x=ln⁡20ln⁡5x=\dfrac{\ln 20}{\ln 5} is perfectly correct; only round (≈1.861\approx 1.861) if the problem asks for a decimal.

Try it
Try it: solve

Solve 9x−1=279^{x - 1} = 27 exactly, and 52x=405^{2x} = 40 to three decimals.

Show the answer

32x−2=333^{2x - 2} = 3^3, so 2x−2=32x - 2 = 3 and x=52x = \tfrac52. For the second, 2xln⁡5=ln⁡402x\ln 5 = \ln 40, so x=ln⁡402ln⁡5≈1.146x = \tfrac{\ln 40}{2\ln 5} \approx 1.146.

Solving Logarithmic Equations

Concept
Strategy & the Domain Check
  • Use log laws to condense into a single log if needed.
  • Rewrite in exponential form, or use “one-to-one” (log⁡bM=log⁡bN⇒M=N\log_b M = \log_b N \Rightarrow M = N).
  • Solve the resulting equation.
  • Check every answer: the input of any log must be positive. Discard extraneous solutions.

Condensing log⁡2x+log⁡2(x−2)\log_2 x + \log_2(x - 2) into log⁡2(x(x−2))\log_2(x(x - 2)) is valid only where both original logs are defined, x>2x > 2. The condensed form is defined on a larger set (for example at x=−2x = -2, where x(x−2)=8>0x(x - 2) = 8 > 0), so it can produce answers the original equation rejects. The domain check undoes that enlargement.

Example
Worked Example: Extraneous Solution

Solve log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2 (x-2) = 3.

log⁡2(x(x−2))=3product rulex(x−2)=23=8exponential formx2−2x−8=0 ⇒ (x−4)(x+2)=0\begin{aligned} \log_2\big(x(x-2)\big) &= 3 &&\text{product rule}\\ x(x-2) &= 2^3 = 8 &&\text{exponential form}\\ x^2 - 2x - 8 &= 0 \ \Rightarrow\ (x-4)(x+2)=0 \end{aligned}

So x=4x = 4 or x=−2x = -2. Check: x=−2x=-2 makes log⁡2(−2)\log_2(-2) undefined — reject it. Only x=4x = 4 works.

Side note
Where it goes wrong

Keeping x=−2x = -2 because it satisfies x(x−2)=8x(x - 2) = 8. In the original equation log⁡2(−2)\log_2(-2) is undefined, so −2-2 is extraneous; the only solution is 44.

Tip

Never skip the check. Condensing can introduce solutions that violate a log's domain. The valid domain here was x>2x>2, which −2-2 fails.

Try it
Try it: solve and check

Solve log⁡3(x+1)+log⁡3(x−1)=1\log_3(x + 1) + \log_3(x - 1) = 1.

Show the answer

Condense: log⁡3(x2−1)=1\log_3(x^2 - 1) = 1, so x2−1=3x^2 - 1 = 3 and x=±2x = \pm 2. The domain is x>1x > 1, so x=−2x = -2 is rejected. Solution: x=2x = 2 (check: log⁡33+log⁡31=1+0=1\log_3 3 + \log_3 1 = 1 + 0 = 1).

Applications

Concept
Models You Will See
  • Growth / decay: A=A0(1±r)tA = A_0(1 \pm r)^t or continuous A=A0ektA = A_0 e^{kt}.
  • Half-life: A=A0(12)t/hA = A_0\left(\tfrac{1}{2}\right)^{t/h}, where hh is the half-life.
  • pH: pH=−log⁡[H+]\text{pH} = -\log[\mathrm{H}^+], where [H+][\mathrm{H}^+] is the hydrogen-ion concentration.
  • Richter magnitude: M=log⁡ ⁣II0M = \log\!\dfrac{I}{I_0}; each whole number is a ×10\times 10 jump in intensity.

Half-life models say that equal times multiply the amount by equal factors: every 88 days, one half. The exponent t8\tfrac t8 counts how many half-lives have passed. When the unknown is the time itself, it sits in the exponent, and a logarithm is the tool that brings it down.

Example
Worked Example: Half-Life
tg8162450100150(24, 25)
Every 88 days the sample halves: 200,100,50,25200, 100, 50, 25.

A 200200 g sample has a half-life of 88 days. How much remains after 2424 days?

A=200(12)24/8=200(12)3=200⋅18=25 g.A = 200\left(\tfrac{1}{2}\right)^{24/8} = 200\left(\tfrac{1}{2}\right)^{3} = 200\cdot\tfrac{1}{8} = 25 \text{ g}.
Side note
Where it goes wrong

Subtracting instead of halving: "200200 g loses 100100 g every 88 days, so after 2424 days nothing is left." Each half-life removes half of what remains: 200→100→50→25200 \to 100 \to 50 \to 25 g.

Example
Worked Example: Solving for Time

A town of 20,00020{,}000 grows continuously at 3%3\% per year. When will it reach 30,00030{,}000?

30000=20000 e0.03t1.5=e0.03tln⁡1.5=0.03t ⇒ t=ln⁡1.50.03≈13.5 years.\begin{aligned} 30000 &= 20000\,e^{0.03t}\\ 1.5 &= e^{0.03t}\\ \ln 1.5 &= 0.03t \ \Rightarrow\ t = \frac{\ln 1.5}{0.03} \approx 13.5 \text{ years.} \end{aligned}
Example
Worked Example: pH

A solution has [H+]=10−4[\mathrm{H}^+] = 10^{-4}. Then pH=−log⁡(10−4)=−(−4)=4\text{pH} = -\log(10^{-4}) = -(-4) = 4. A pH below 77 is acidic.

Tip

Big picture: whenever the unknown is in the exponent (“how long?”, “what rate?”), a logarithm is the tool that frees it.

Try it
Try it: doubling time

A culture grows continuously at 12%12\% per hour. How long until it doubles? How long to go from 500500 to 80008000?

Show the answer

Doubling: 2=e0.12t2 = e^{0.12t}, so t=ln⁡20.12≈5.78t = \tfrac{\ln 2}{0.12} \approx 5.78 hours. From 500500 to 80008000 is a factor of 16=2416 = 2^4, four doublings: about 4×5.78≈23.14 \times 5.78 \approx 23.1 hours.

Going Deeper: Advanced Exponential & Log Ideas

Concept
Change-of-Base Chains & Telescoping Products

Change of base can be written in any common base cc:

log⁡bM=log⁡cMlog⁡cb.\log_b M = \frac{\log_c M}{\log_c b}.

This is the key to telescoping a product of logs. Watch a chain of log⁡k(k+1)\log_k(k{+}1) terms collapse when every log is rewritten over the same base cc:

log⁡23⋅log⁡34⋅log⁡45⋯log⁡n−1n=log⁡3log⁡2⋅log⁡4log⁡3⋅log⁡5log⁡4⋯log⁡nlog⁡(n−1).\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{n-1} n = \frac{\log 3}{\log 2}\cdot\frac{\log 4}{\log 3}\cdot\frac{\log 5}{\log 4}\cdots\frac{\log n}{\log(n-1)}.

Each numerator cancels the next denominator, leaving log⁡nlog⁡2=log⁡2n\dfrac{\log n}{\log 2} = \log_2 n. The whole chain equals a single log.

Example
Worked Example: A Telescoping Log Product

Evaluate log⁡23⋅log⁡34⋅log⁡45⋅log⁡56⋅log⁡67⋅log⁡78\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdot \log_5 6 \cdot \log_6 7 \cdot \log_7 8.

∏=log⁡3log⁡2⋅log⁡4log⁡3⋅log⁡5log⁡4⋅log⁡6log⁡5⋅log⁡7log⁡6⋅log⁡8log⁡7change every log to base 10=log⁡8log⁡2=log⁡28=3everything cancels but the ends\begin{aligned} \prod &= \frac{\log 3}{\log 2}\cdot\frac{\log 4}{\log 3}\cdot\frac{\log 5}{\log 4}\cdot\frac{\log 6}{\log 5}\cdot\frac{\log 7}{\log 6}\cdot\frac{\log 8}{\log 7} &&\text{change every log to base }10\\ &= \frac{\log 8}{\log 2} = \log_2 8 = 3 &&\text{everything cancels but the ends} \end{aligned}

The product of six “ugly” logs is exactly 33.

Concept
“Let log⁡23=a\log_2 3 = a” — Expressing Logs Symbolically

When a problem fixes one log as a letter, express others by breaking numbers into prime factors and applying the three laws. If log⁡23=a\log_2 3 = a, then:

  • log⁡26=log⁡2(2⋅3)=1+a\log_2 6 = \log_2(2\cdot 3) = 1 + a;
  • log⁡212=log⁡2(22⋅3)=2+a\log_2 12 = \log_2(2^2\cdot 3) = 2 + a;
  • log⁡298=log⁡232−log⁡223=2a−3\log_2 \tfrac{9}{8} = \log_2 3^2 - \log_2 2^3 = 2a - 3;
  • log⁡32=1log⁡23=1a\log_3 2 = \dfrac{1}{\log_2 3} = \dfrac{1}{a}  (reciprocal / change of base).
Example
Worked Example: Express log⁡648\log_6 48 in terms of a=log⁡23a=\log_2 3

Change to base 22, then factor 48=24⋅348 = 2^4\cdot 3 and 6=2⋅36 = 2\cdot 3:

log⁡648=log⁡248log⁡26change of base=log⁡2(24⋅3)log⁡2(2⋅3)=4+log⁡231+log⁡23product + power laws=4+a1+asubstitute a=log⁡23\begin{aligned} \log_6 48 &= \frac{\log_2 48}{\log_2 6} &&\text{change of base}\\ &= \frac{\log_2(2^4\cdot 3)}{\log_2(2\cdot 3)} = \frac{4 + \log_2 3}{1 + \log_2 3} &&\text{product + power laws}\\ &= \frac{4 + a}{1 + a} &&\text{substitute } a=\log_2 3 \end{aligned}
Concept
Exponential Equations Hidden as Quadratics

An equation with terms in 2x2^x and 22x2^{2x} (or exe^x and e2xe^{2x}) is secretly a quadratic. Substitute u=2x>0u = 2^x > 0, using 22x=(2x)2=u22^{2x} = (2^x)^2 = u^2, solve the quadratic, then back-substitute. Reject any u≤0u \le 0, since 2x2^x is always positive.

Example
Worked Example: Reducible to a Quadratic

Solve 4x−5⋅2x+4=04^x - 5\cdot 2^x + 4 = 0. Since 4x=(2x)24^x = (2^x)^2, let u=2xu = 2^x:

u2−5u+4=0substitute u=2x(u−1)(u−4)=0 ⇒ u=1 or u=4factor2x=1 ⇒ x=0,2x=4 ⇒ x=2back-substitute\begin{aligned} u^2 - 5u + 4 &= 0 &&\text{substitute } u=2^x\\ (u-1)(u-4) &= 0 \ \Rightarrow\ u = 1 \text{ or } u = 4 &&\text{factor}\\ 2^x = 1 \ \Rightarrow\ x = 0, \qquad 2^x &= 4 \ \Rightarrow\ x = 2 &&\text{back-substitute} \end{aligned}

Both uu-values are positive, so both solutions are valid: x=0x = 0 and x=2x = 2.

Concept
The xlog⁡xx^{\log x} Type — Take a Log of a Log-Exponent

When the variable appears in both the base and the exponent, as in xlog⁡xx^{\log x}, take log⁡\log of both sides first; the power rule turns the exponent into a factor and creates a quadratic in log⁡x\log x. Let t=log⁡xt = \log x.

Example
Worked Example: Solving xlog⁡x=100xx^{\log x} = 100x

Take log⁡10\log_{10} of both sides and let t=log⁡xt = \log x:

log⁡ ⁣(xlog⁡x)=log⁡(100x)(log⁡x)(log⁡x)=log⁡100+log⁡xpower & product rulest2=2+t ⇒ t2−t−2=0substitute t=log⁡x(t−2)(t+1)=0 ⇒ t=2 or t=−1factorlog⁡x=2 ⇒ x=100,log⁡x=−1 ⇒ x=110\begin{aligned} \log\!\left(x^{\log x}\right) &= \log(100x)\\ (\log x)(\log x) &= \log 100 + \log x &&\text{power \& product rules}\\ t^2 &= 2 + t \ \Rightarrow\ t^2 - t - 2 = 0 &&\text{substitute } t=\log x\\ (t-2)(t+1) &= 0 \ \Rightarrow\ t = 2 \text{ or } t = -1 &&\text{factor}\\ \log x = 2 \ \Rightarrow\ x = 100, &\qquad \log x = -1 \ \Rightarrow\ x = \tfrac{1}{10} \end{aligned}

Both are positive, so x=100x = 100 and x=110x = \tfrac{1}{10} both check.

Concept
Systems of Logarithmic Equations

A system mixing sums and products of logs is solved by condensing each equation, then converting to ordinary algebra. Keep the same base throughout, and enforce the domain (every argument >0>0) at the end.

Example
Worked Example: A Log System

Solve  log⁡2x+log⁡2y=5\ \log_2 x + \log_2 y = 5 and  log⁡2x−log⁡2y=1\ \log_2 x - \log_2 y = 1.

log⁡2(xy)=5 ⇒ xy=25=32product rulelog⁡2 ⁣xy=1 ⇒ xy=21=2quotient rule\begin{aligned} \log_2(xy) &= 5 \ \Rightarrow\ xy = 2^5 = 32 &&\text{product rule}\\ \log_2\!\tfrac{x}{y} &= 1 \ \Rightarrow\ \tfrac{x}{y} = 2^1 = 2 &&\text{quotient rule} \end{aligned}

So x=2yx = 2y, giving 2y2=322y^2 = 32, hence y=4y = 4 (reject y=−4y=-4, out of domain) and x=8x = 8. Check: log⁡28+log⁡24=3+2=5\log_2 8 + \log_2 4 = 3 + 2 = 5, as required.

Concept
Continuous Compounding & Logistic Growth

Two continuous models worth knowing beyond A=PertA = Pe^{rt}:

  • Doubling / tripling time: set A=2PA = 2P in PertPe^{rt} to get t=ln⁡2rt = \dfrac{\ln 2}{r} (the “rule of 6969” in disguise).
  • Logistic growth: real populations cannot grow forever; they level off at a carrying capacity LL: @@BLOCK0@@ Early on it looks exponential; as t→∞t\to\infty, e−kt→0e^{-kt}\to 0 so P→LP\to L. The curve is S-shaped (sigmoidal) with a horizontal asymptote at P=LP = L.

Logistic growth rises like an exponential, then bends over and approaches the carrying capacity LL.

Tip

Continuous vs. logistic: pure PertPe^{rt} has no ceiling and grows without bound; logistic growth throttles itself as PP nears LL. Use logistic whenever a resource limit (space, food, market size) caps the total.

Concept
Logarithmic & Exponential Inequalities

Solving an inequality adds a direction rule on top of the equation techniques:

  • Because log⁡bx\log_b x (with b>1b>1) is increasing, applying it preserves the inequality: M<N⇔log⁡bM<log⁡bNM < N \Leftrightarrow \log_b M < \log_b N for M,N>0M,N>0.
  • If the base satisfies 0<b<10 < b < 1, both bxb^x and log⁡bx\log_b x are decreasing, so the inequality flips.
  • Always intersect your answer with the domain (arguments of every log must be positive).
Example
Worked Example: A Log Inequality

Solve log⁡2(x−1)<3\log_2(x - 1) < 3. First the domain: x−1>0x - 1 > 0, so x>1x > 1. Since base 2>12 > 1 is increasing, rewrite 3=log⁡283 = \log_2 8 and drop the logs:

x−1<23=8 ⇒ x<9.x - 1 < 2^3 = 8 \ \Rightarrow\ x < 9.

Intersect with the domain x>1x > 1: the solution is 1<x<91 < x < 9.

Tip

Two-part discipline for inequalities: (1) solve the associated equation for the boundary, (2) decide the direction from whether the base is >1>1 (keep) or <1<1 (flip), then (3) trim to the domain. Missing the domain step is the #1 error.

Why It Works

Let x=bmx = b^m and y=bny = b^n, so m=log⁡bxm = \log_bx, n=log⁡byn = \log_by. Then xy=bm+nxy = b^{m+n}, so log⁡b(xy)=m+n=log⁡bx+log⁡by\log_b(xy) = m + n = \log_bx + \log_by: the product law is the exponent law bmbn=bm+nb^mb^n = b^{m+n} in logarithmic clothing. The power law comes from (bm)k=bmk(b^m)^k = b^{mk} the same way. Change of base: if bc=ab^c = a, take ln⁡\ln of both sides — cln⁡b=ln⁡ac\ln b = \ln a, so c=ln⁡aln⁡bc = \tfrac{\ln a}{\ln b}.

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
Logarithm rules
log⁡b(xy)=log⁡bx+log⁡by,log⁡b ⁣xy=log⁡bx−log⁡by,log⁡b(xn)=nlog⁡bx\log_b(xy)=\log_b x+\log_b y,\quad \log_b\!\frac{x}{y}=\log_b x-\log_b y,\quad \log_b(x^{n})=n\log_b x

What it means. A logarithm answers "what exponent?", so it turns multiplication into addition.

Example. log⁡240−log⁡25=log⁡2405=log⁡28=3\log_2 40-\log_2 5=\log_2\tfrac{40}{5}=\log_2 8=3.

Why it works. Let x=bmx=b^{m} and y=bny=b^{n}, so log⁡bx=m\log_b x=m and log⁡by=n\log_b y=n. Then xy=bmbn=bm+nxy=b^{m}b^{n}=b^{m+n}, whose logarithm is m+nm+n — exactly log⁡bx+log⁡by\log_b x+\log_b y. The other rules follow the same way from the exponent rules.

Tip. log⁡(x+y)\log(x+y) does not simplify. The rules only apply to products, quotients and powers inside the log.

Tip
Change of base
log⁡bx=log⁡kxlog⁡kb\log_b x = \frac{\log_k x}{\log_k b}

What it means. Rewrite any logarithm in a base your calculator knows.

Example. log⁡210=log⁡10log⁡2≈10.301≈3.32\log_2 10=\dfrac{\log 10}{\log 2}\approx\dfrac{1}{0.301}\approx 3.32.

Why it works. Let y=log⁡bxy=\log_b x, so by=xb^{y}=x. Take log⁡k\log_k of both sides: ylog⁡kb=log⁡kxy\log_k b=\log_k x, then divide.

Tip. Use base 1010 or base ee — both are on every calculator.