Honors Pre-Calculus Unit 4 Review: Exponential and Logarithmic Functions

An exponential function has the form $f(x) = ab^x$, where $a$ is the initial value and $b$ is the base, with $b > 0$ and $b \neq 1$. Every time $x$ goes up by 1, the output gets multiplied by $b$. A linear function adds the same amount each step instead.

For a positive initial value $a$, if $b > 1$ you get growth, and if $0 < b < 1$ you get decay. A percent rate turns into a factor by adding or subtracting from 1. So 5% growth means $b = 1.05$, and 15% decay means $b = 0.85$.

The number $e \approx 2.71828$ is the natural base. It shows up in continuous growth, written $A = A_0e^{kt}$, where $k$ is the continuous rate.

Compound interest uses $A = P\left(1 + \frac{r}{n}\right)^{nt}$, where $n$ is the number of compounding periods per year. Continuous compounding uses $A = Pe^{rt}$.

Worked example. Invest 1,000 dollars at 6% compounded monthly for 5 years. Here $r = 0.06$, $n = 12$ and $t = 5$, so $A = 1000\left(1 + \frac{0.06}{12}\right)^{60} = 1000(1.005)^{60} \approx 1348.85$ dollars. With continuous compounding, $A = 1000e^{0.3} \approx 1349.86$ dollars. Compounding more often helps, but here only by about a dollar.

The transformed form is $f(x) = a \cdot b^{x-h} + k$, where $a \neq 0$, $b > 0$ and $b \neq 1$. Read the parameters one at a time.

The horizontal asymptote is $y = k$. For the parent $y = b^x$, that's the x-axis, which is why $b^x$ never equals zero or goes negative. The range is $(k, \infty)$ when $a > 0$ and $(-\infty, k)$ when $a < 0$.

The value $h$ shifts the graph right $h$ units, so $2^{x+3}$ moves left 3. A common slip is assuming $(0, a)$ is still the y-intercept after a shift. Find the actual y-intercept by plugging in $x = 0$.

For growth with $a > 0$, the graph rises without bound as $x \to \infty$ and flattens toward $y = k$ as $x \to -\infty$. Decay reverses that. A larger base for growth, or a base closer to 0 for decay, makes the curve steeper.

Watch the two reflections. Changing the sign of $a$ reflects the graph across $y = k$. For example, $-2^x$ means $-(2^x)$ and is the reflection of $2^x$ across the x-axis. Replacing $x$ with $-x$ flips the graph over the y-axis, and since $b^{-x} = \left(\frac{1}{b}\right)^x$, that turns growth into decay.

A logarithm is an exponent. The statement $\log_b(x) = y$ means exactly the same thing as $b^y = x$. So $\log_2(32) = 5$ because $2^5 = 32$.

When you're unsure, rewrite in exponential form and ask what power of the base gives the argument. For $\log_9(3)$, you need $9^y = 3$, so $y = \frac{1}{2}$. For $\log(0.01)$, the base is 10 and $10^{-2} = 0.01$, so the answer is $-2$. A log's output can be negative or a fraction. The argument is the part that has to be positive.

Two bases get special notation. The common log $\log(x)$ has base 10, and the natural log $\ln(x)$ has base $e$.

Since $\log_b(x)$ and $b^x$ are inverse functions, they cancel each other.

• $\log_b(b^x) = x$, so $\ln(e^4) = 4$
• $b^{\log_b(x)} = x$, so $10^{\log 7} = 7$

Two values come up constantly. $\log_b(1) = 0$ because $b^0 = 1$, and $\log_b(b) = 1$ because $b^1 = b$. You can't take the log of 0 or a negative number, because no power of a positive base produces those outputs.

The graph of $y = \log_b(x)$ is the graph of $y = b^x$ reflected over the line $y = x$. Every feature swaps. The exponential's horizontal asymptote $y = 0$ becomes a vertical asymptote $x = 0$, and its y-intercept $(0, 1)$ becomes an x-intercept $(1, 0)$.

For $f(x) = a\log_b(x - h) + k$, where $a \neq 0$, $b > 0$ and $b \neq 1$, find the domain by setting the argument greater than zero. That gives $x > h$, and the vertical asymptote sits at $x = h$. The range is all real numbers.

For $b > 1$, the parent graph $y = \log_b(x)$ increases but bends over, rising more and more slowly. It still keeps going up forever. A log graph has no horizontal asymptote, which is a frequent mix-up.

Worked example. Find the features of $f(x) = \log_3(x + 2) - 1$. Setting $x + 2 > 0$ gives the domain $x > -2$ and the asymptote $x = -2$. For the x-intercept, set $\log_3(x + 2) = 1$, so $x + 2 = 3$ and $x = 1$. For the y-intercept, $f(0) = \log_3(2) - 1 \approx -0.37$. Another easy point is $x = 7$, where $f(7) = \log_3(9) - 1 = 1$.

Three rules let you rewrite logs, and each one is an exponent rule in disguise. In the rules below, $b > 0$, $b \neq 1$, $M > 0$ and $N > 0$.

The product rule says $\log_b(MN) = \log_b(M) + \log_b(N)$, matching $b^m \cdot b^n = b^{m+n}$. The quotient rule says $\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$. The power rule says $\log_b(M^n) = n\log_b(M)$, which is how you pull a variable out of an exponent.

Expanding breaks one log into several. Condensing runs the rules backward to make a single log, and it only works when every log has the same base.

Worked example. Expand $\log_3\left(\frac{9x^2}{y}\right)$. The original expression requires $x \neq 0$ and $y > 0$. Since $x^2 = |x|^2$ and $|x| > 0$, the log properties give $2 + 2\log_3(|x|) - \log_3(y)$.

There is no rule for the log of a sum, so $\log(x + 5)$ stays as it is.

To evaluate a log with an awkward base, use the change-of-base formula $\log_b(x) = \frac{\ln x}{\ln b}$. For example, $\log_3(5) = \frac{\ln 5}{\ln 3} \approx 1.465$.

Pick a method based on what the equation looks like.

If both sides can be written with the same base, set the exponents equal. For $4^x = 8$, rewrite as $2^{2x} = 2^3$, so $2x = 3$ and $x = 1.5$.

If the bases won't match, isolate the exponential, then take $\ln$ of both sides and use the power rule to bring the exponent down.

Worked example. Solve $5 \cdot 3^x = 200$. Divide by 5 first to get $3^x = 40$. Taking the natural log gives $x\ln 3 = \ln 40$, so $x = \frac{\ln 40}{\ln 3} \approx 3.358$. Don't multiply $5 \cdot 3^x$ into $15^x$, since the exponent applies only to the 3.

For log equations, condense into a single log, then convert to exponential form. If both sides are single logs with the same base, set the arguments equal instead.

Worked example. Solve $\log_2(x) + \log_2(x - 2) = 3$. Condense to $\log_2\big(x(x - 2)\big) = 3$, so $x^2 - 2x = 8$. Factoring $(x - 4)(x + 2) = 0$ gives $x = 4$ or $x = -2$. Plugging in $-2$ produces $\log_2(-2)$, which is undefined, so the only solution is $x = 4$.

Most models in this unit are versions of $y = ab^t$ with a story attached.

Percent growth or decay uses $y = a(1 \pm r)^t$. Half-life uses $A = A_0\left(\frac{1}{2}\right)^{t/h}$, where $h$ is the half-life, and doubling time works the same way with base 2. Continuous models use $A = A_0e^{kt}$. You can switch between forms with $b = e^k$, or $k = \ln b$, so 5% yearly growth is a continuous rate of $k = \ln 1.05 \approx 0.0488$.

Newton's Law of Cooling, $T(t) = T_s + (T_0 - T_s)e^{-kt}$, models an exponentially shrinking temperature gap when the surrounding temperature $T_s$ is constant and $k > 0$. In the logistic model $P(t) = \frac{K}{1 + Ae^{-rt}}$, positive values of $K$, $A$ and $r$ give an S-shaped curve that approaches the carrying capacity $K$ from below.

Worked example. Coffee at 190°F sits in a 70°F room and is 150°F after 10 minutes. The gap starts at 120 degrees and drops to 80, so $e^{-10k} = \frac{2}{3}$ and $k = \frac{\ln 1.5}{10} \approx 0.0405$. To reach 100°F, the gap must be 30, so $e^{-kt} = \frac{1}{4}$ and $t = \frac{\ln 4}{k} \approx 34.2$ minutes.

Before fitting anything, check whether an exponential model makes sense. For equally spaced x-values, linear data has constant differences and exponential data has roughly constant ratios.

Worked example. A culture measures 200, 260, 338 and 439.4 at hours 0 through 3. The ratios are $\frac{260}{200} = 1.3$, $\frac{338}{260} = 1.3$ and $\frac{439.4}{338} = 1.3$. So $a = 200$, $b = 1.3$, and the model is $y = 200(1.3)^x$, meaning 30% growth per hour.

With two points that aren't at $x = 0$, divide one equation by the other to cancel $a$. From $(2, 50)$ and $(5, 400)$ you get $b^3 = 8$, so $b = 2$. Then $50 = a \cdot 2^2$ gives $a = 12.5$.

Real data is messy, so exponential regression estimates a model $y = ab^x$. One method fits a line to $(x, \ln y)$ for positive $y$-values, because taking logs turns $y = ab^x$ into $\ln y = \ln a + x\ln b$. Other methods fit the exponential curve using errors on the original scale, so their estimates can differ. An exact positive exponential model appears as a straight line on a semi-log plot.

Judge the fit with a residual plot. Random scatter around zero supports the model, but does not prove it fits. A curved pattern suggests the model may be missing a relationship in the data. Be careful predicting far past your data, since real growth usually slows down.