Honors Algebra II Unit 8 Review: Exponential and Logarithmic Functions

A logarithm is an exponent. The statement $\log_b y = x$ means the same thing as $b^x = y$. So $\log_2 8 = 3$ because $2^3 = 8$, and $\log 0.01 = -2$ because $10^{-2} = 0.01$. The base $b$ has to be positive and not equal to 1.

Two bases get their own notation. The common log, $\log x$, uses base 10. The natural log, $\ln x$, uses base $e \approx 2.718$.

Because $y = \log_b x$ is the inverse of $y = b^x$, its graph is the exponential graph reflected over the line $y = x$. The domain and range swap. The log has domain $x > 0$, range all real numbers, a vertical asymptote at $x = 0$, and an x-intercept at $(1, 0)$.

The log properties are exponent rules translated into log form. In these rules, $M$ and $N$ must be positive. Multiplying powers adds exponents, so $\log_b(MN) = \log_b M + \log_b N$. Dividing subtracts them, so $\log_b \frac{M}{N} = \log_b M - \log_b N$. A power inside the log moves out front, so $\log_b(M^n) = n\log_b M$.

For a base your calculator doesn't have, use change of base. For example, $\log_3 50 = \frac{\ln 50}{\ln 3} \approx \frac{3.912}{1.099} \approx 3.56$.

For many exponential equations, isolate a power and then take a log of both sides. For many log equations, condense the log expressions and rewrite the equation in exponential form. Other equations may need different methods.

If both sides can be written with the same base, you don't need logs. For $4^{x+1} = 8^x$, rewrite both sides in base 2 to get $2^{2x+2} = 2^{3x}$. Then $2x + 2 = 3x$, so $x = 2$.

Worked example. Solve $5 \cdot 3^x + 2 = 102$. Subtract 2 to get $5 \cdot 3^x = 100$, then divide by 5 to get $3^x = 20$. Take ln of both sides to get $x \ln 3 = \ln 20$. So $x = \frac{\ln 20}{\ln 3} \approx \frac{2.996}{1.099} \approx 2.73$.

Worked example. Solve $\log_2 x + \log_2(x - 2) = 3$. The product rule condenses this to $\log_2[x(x - 2)] = 3$. In exponential form that is $x(x - 2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$. Factoring gives $(x - 4)(x + 2) = 0$, so $x = 4$ or $x = -2$. Substituting $x = -2$ into the original equation gives $\log_2(-2)$, which is undefined, so $-2$ is extraneous. The only solution is $x = 4$.

Whenever you condense logs, check every answer in the original equation, not the condensed one.

An exponential function has the form $f(x) = a \cdot b^x$ with $a \neq 0$, $b > 0$, and $b \neq 1$. The value $a$ is the y-intercept, since $f(0) = a \cdot b^0 = a$. The base $b$ is the factor the output gets multiplied by each time $x$ goes up by 1.

That constant multiplier is what separates exponential functions from linear ones. A linear function adds the same amount each step, while an exponential function multiplies by the same factor. When $a > 0$, the function shows growth if $b > 1$ and decay if $0 < b < 1$. A negative $a$ reflects the graph across the x-axis. A percent rate $r$ becomes a factor of $b = 1 + r$ for growth or $b = 1 - r$ for decay, so a 15% drop means $b = 0.85$.

With $a > 0$, the graph of $f(x) = a \cdot b^x$ has domain all real numbers, range $y > 0$, and a horizontal asymptote at $y = 0$. That holds for growth and for decay. In $f(x) = a \cdot b^{x - h} + k$, the graph shifts right $h$ units and up $k$ units, and the asymptote moves to $y = k$.

The exponent rules apply here. They are $b^m \cdot b^n = b^{m+n}$, $\frac{b^m}{b^n} = b^{m-n}$, $(b^m)^n = b^{mn}$, $b^{-n} = \frac{1}{b^n}$, and $b^{m/n} = \sqrt[n]{b^m}$. For example, $27^{2/3} = (\sqrt[3]{27})^2 = 9$.

The number $e \approx 2.71828$ is the base that comes up in continuous growth. The graph of $f(x) = e^x$ sits between $2^x$ and $3^x$.

Most applications use one of a few models. Growth or decay by a fixed percent per period uses $A = a \cdot b^t$. Compound interest uses $A = P\left(1 + \frac{r}{n}\right)^{nt}$. Here $P$ is the starting amount, $r$ is the annual rate as a decimal, $n$ is the number of compounding periods per year, and $t$ is time in years. Continuous growth uses $A = Pe^{rt}$, and half-life uses $A = A_0\left(\frac{1}{2}\right)^{t/h}$, where $h$ is the half-life.

For example, 1000 dollars at 5% compounded monthly for 10 years grows to $1000\left(1 + \frac{0.05}{12}\right)^{120} \approx 1647.01$ dollars. Compounded continuously, the same deposit grows to $1000e^{0.5} \approx 1648.72$ dollars.

Worked example. A medicine has a half-life of 6 hours, and a patient gets an 80 mg dose. When is 15 mg left? Set up $80\left(\frac{1}{2}\right)^{t/6} = 15$. Dividing by 80 gives $\left(\frac{1}{2}\right)^{t/6} = 0.1875$. Taking ln of both sides gives $\frac{t}{6}\ln 0.5 = \ln 0.1875$. So $t = \frac{6\ln 0.1875}{\ln 0.5} \approx \frac{6(-1.674)}{-0.693} \approx 14.5$ hours.

Logs also build scales for values that span huge ranges. Using $\text{pH} \approx -\log[H^+]$, where $[H^+]$ is the hydrogen ion concentration in moles per liter, a drop from pH 5 to pH 3 means about 100 times the hydrogen ion concentration. In decibels, $L = 10\log\frac{I}{I_0}$, so every 10 dB increase means 10 times the intensity.