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,a0,  b>0,  b1.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.

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  • 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.

Growth y=2xy=2^x (red) rises to the right; decay y=(12)xy=\left(\tfrac12\right)^x (blue) 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=abxh+ky = a\,b^{\,x-h} + k is:

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  • 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=2x2y = 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 202=12^0 - 2 = -1.

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=34xy = 3\cdot 4^x.

The Number ee & Continuous Growth

Concept
Euler's Number ee

The constant e2.71828e \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 nn \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.

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

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

monthly: A=1000(1+0.0612)125=1000(1.005)60$1348.85,continuous: A=1000e0.065=1000e0.3$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 nn\to\infty.

Tip

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

Logarithms: Definition & Evaluating

Concept
A Logarithm is an Exponent

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

logby=xbx=y,b>0, b1, y>0.\log_b y = x \quad\Longleftrightarrow\quad b^x = y, \qquad b>0,\ b\neq 1,\ y>0.

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

Example
Worked Example: Evaluating
log232=5because 25=32,log525=2because 52=25,log84=23because 82/3=(23)2/3=22=4,logb1=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}
Concept
Common and Natural Logs

Two bases are so useful they get their own notation:

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  • Common log: logx\log x means log10x\log_{10} x (base 10).
  • Natural log: lnx\ln x means logex\log_e x (base ee).

Inverse facts: log10k=k\log 10^k = k, lnek=k\ln e^k = k, 10logx=x10^{\log x}=x, elnx=xe^{\ln x}=x.

Tip

Tip: log1000=3\log 1000 = 3 and lne2=2\ln e^2 = 2 take no calculator --- read them straight from the definition.

Graphs of Logarithmic Functions

Concept
The Inverse of an Exponential

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

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  • Domain x>0x > 0; range all real numbers.
  • Vertical asymptote at x=0x = 0 (the yy-axis).
  • passes through (1,0)(1,0) since logb1=0\log_b 1 = 0.

A shift y=logb(xh)y = \log_b(x-h) moves the asymptote to x=hx = h and the domain to x>hx > h.

y=log2xy=\log_2 x (blue) is the mirror image of y=2xy=2^x (red) across y=xy=x. Its vertical asymptote is the yy-axis, x=0x=0.

Example
Worked Example: Domain of a Shifted Log

For y=log2(x3)y = \log_2(x - 3), the inside must be positive: x3>0x - 3 > 0, so the domain is x>3x > 3 and the vertical asymptote is x=3x = 3.

Properties of Logarithms

Concept
The Three Laws (same base bb)
Product:logb(MN)=logbM+logbN,Quotient:logb ⁣(MN)=logbMlogbN,Power:logb(Mp)=plogbM.\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.

Concept
Change of Base

To evaluate a log in any base with a calculator:

logbM=logMlogb=lnMlnb.\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.

logx3yz2=log(x3y)logz2quotient=logx3+logy1/2logz2product=3logx+12logy2logzpower\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}
Example
Worked Example: Condensing

Condense 2lnx+lnyln32\ln x + \ln y - \ln 3 into a single logarithm.

2lnx+lnyln3=lnx2+lnyln3power=ln(x2y)ln3product=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 logMlogNlogMN\dfrac{\log M}{\log N} \neq \log\dfrac{M}{N}.

Solving Exponential Equations

Concept
Two Strategies
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  • 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)=nxn1,(fg)=fg+fg,(fg)=fgfgg2,(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)
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}.
Example
Worked Example: Taking a Log

Solve 32x=503\cdot 2^{x} = 50. Isolate the power, then take ln\ln of both sides.

2x=503ln ⁣(2x)=ln ⁣503take the natural logxln2=ln ⁣503power rule brings x downx=ln(50/3)ln2=log2 ⁣5034.059exact, 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=ln20ln5x=\dfrac{\ln 20}{\ln 5} is perfectly correct; only round (1.861\approx 1.861) if the problem asks for a decimal.

Solving Logarithmic Equations

Concept
Strategy & the Domain Check
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  • Use log laws to condense into a single log if needed.
  • Rewrite in exponential form, or use “one-to-one” (logbM=logbNM=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.
Example
Worked Example: Extraneous Solution

Solve log2x+log2(x2)=3\log_2 x + \log_2 (x-2) = 3.

log2(x(x2))=3product rulex(x2)=23=8exponential formx22x8=0  (x4)(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 log2(2)\log_2(-2) undefined --- reject it. Only x=4x = 4 works.

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.

Applications

Concept
Models You Will See
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  • 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.
Example
Worked Example: Half-Life

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=20018=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}.
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=20000e0.03t1.5=e0.03tln1.5=0.03t  t=ln1.50.0313.5 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+]=104[\mathrm{H}^+] = 10^{-4}. Then pH=log(104)=(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.

Going Deeper: Advanced Exponential & Log Ideas

Concept
Change-of-Base Chains & Telescoping Products

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

logbM=logcMlogcb.\log_b M = \frac{\log_c M}{\log_c b}.

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

log23log34log45logn1n=log3log2log4log3log5log4lognlog(n1).\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 lognlog2=log2n\dfrac{\log n}{\log 2} = \log_2 n. The whole chain equals a single log.

Example
Worked Example: A Telescoping Log Product

Evaluate log23log34log45log56log67log78\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdot \log_5 6 \cdot \log_6 7 \cdot \log_7 8.

=log3log2log4log3log5log4log6log5log7log6log8log7change every log to base 10=log8log2=log28=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 log23=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 log23=a\log_2 3 = a, then:

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  • log26=log2(23)=1+a\log_2 6 = \log_2(2\cdot 3) = 1 + a;
  • log212=log2(223)=2+a\log_2 12 = \log_2(2^2\cdot 3) = 2 + a;
  • log298=log232log223=2a3\log_2 \tfrac{9}{8} = \log_2 3^2 - \log_2 2^3 = 2a - 3;
  • log32=1log23=1a\log_3 2 = \dfrac{1}{\log_2 3} = \dfrac{1}{a}  (reciprocal / change of base).
Example
Worked Example: Express log648\log_6 48 in terms of a=log23a=\log_2 3

Change to base 22, then factor 48=24348 = 2^4\cdot 3 and 6=236 = 2\cdot 3:

log648=log248log26change of base=log2(243)log2(23)=4+log231+log23product + power laws=4+a1+asubstitute a=log23\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 u0u \le 0, since 2x2^x is always positive.

Example
Worked Example: Reducible to a Quadratic

Solve 4x52x+4=04^x - 5\cdot 2^x + 4 = 0. Since 4x=(2x)24^x = (2^x)^2, let u=2xu = 2^x:

u25u+4=0substitute u=2x(u1)(u4)=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 xlogxx^{\log x} Type --- Take a Log of a Log-Exponent

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

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

Take log10\log_{10} of both sides and let t=logxt = \log x:

log ⁣(xlogx)=log(100x)(logx)(logx)=log100+logxpower & product rulest2=2+t  t2t2=0substitute t=logx(t2)(t+1)=0  t=2 or t=1factorlogx=2  x=100,logx=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  log2x+log2y=5\ \log_2 x + \log_2 y = 5 and  log2xlog2y=1\ \log_2 x - \log_2 y = 1.

log2(xy)=5  xy=25=32product rulelog2 ⁣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: log28+log24=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}:

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  • Doubling / tripling time: set A=2PA = 2P in PertPe^{rt} to get t=ln2rt = \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 tt\to\infty, ekt0e^{-kt}\to 0 so PLP\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:

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  • Because logbx\log_b x (with b>1b>1) is increasing, applying it preserves the inequality: M<NlogbM<logbNM < 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 logbx\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 log2(x1)<3\log_2(x - 1) < 3. First the domain: x1>0x - 1 > 0, so x>1x > 1. Since base 2>12 > 1 is increasing, rewrite 3=log283 = \log_2 8 and drop the logs:

x1<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.

Formulas, Proofs & Tips

Tip
Logarithm rules
logb(xy)=logbx+logby,logb ⁣xy=logbxlogby,logb(xn)=nlogbx\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. log240log25=log2405=log28=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 logbx=m\log_b x=m and logby=n\log_b y=n. Then xy=bmbn=bm+nxy=b^{m}b^{n}=b^{m+n}, whose logarithm is m+nm+n — exactly logbx+logby\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
logbx=logkxlogkb\log_b x = \frac{\log_k x}{\log_k b}

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

Example. log210=log10log210.3013.32\log_2 10=\dfrac{\log 10}{\log 2}\approx\dfrac{1}{0.301}\approx 3.32.

Why it works. Let y=logbxy=\log_b x, so by=xb^{y}=x. Take logk\log_k of both sides: ylogkb=logkxy\log_k b=\log_k x, then divide.

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