Unit 11 · Checkpoint quiz

Unit 11 checkpoint: Exponentials & logarithms

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next. Use a calculator only on the two questions that say so.

If $f(x) = 3 \cdot 2^x$, find $f(-2)$.

$f(-2) = 3 \cdot 2^{-2}$. The exponent applies only to the 2, not to the 3.

A negative exponent means the reciprocal: $2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$.

$3 \cdot \dfrac{1}{4} = \dfrac{3}{4}$. Answer: $\dfrac{3}{4}$.

You invest $2,000 at 5% annual interest, compounded monthly. Use $A = P\left(1 + \dfrac{r}{n}\right)^{nt}$ to find the amount after 3 years. Use a calculator and round to the nearest cent.

$P = 2000$, $r = 0.05$, $n = 12$ (monthly), $t = 3$, so $nt = 36$.

$A = 2000\left(1 + \dfrac{0.05}{12}\right)^{36}$. Type it in one go, with parentheses: 2000*(1+0.05/12)^36.

$A \approx 2322.944\ldots$. Answer: $2,322.94.

Evaluate $\log_3 \dfrac{1}{9}$.

Ask: 3 to what power gives $\dfrac{1}{9}$?

$3^2 = 9$, so $3^{-2} = \dfrac{1}{9}$.

Answer: $-2$.

Find the domain of $f(x) = \log(2x - 6)$. Write it in interval notation.

You can only take the log of a positive number: $2x - 6 \gt 0$.

$2x \gt 6$, so $x \gt 3$. (Not $\ge$: $\log 0$ is undefined.)

Answer: $(3, \infty)$.

Expand $\log\left(\dfrac{x^3 y}{100}\right)$. Assume $x$ and $y$ are positive.

$3\log x + \log y + 2$
$3\log x \cdot \log y - 2$
$3\log x + \log y - 2$
$3\log x + \log y - 100$

Quotient rule: $\log(x^3 y) - \log 100$. Product rule: $\log x^3 + \log y - \log 100$.

Power rule: $\log x^3 = 3\log x$. And $\log 100 = 2$ because $10^2 = 100$.

Answer: $3\log x + \log y - 2$.

Write as a single logarithm: $2\ln x - \ln(x + 1) + \ln 3$.

$\ln \dfrac{3x^2}{x + 1}$
$\ln \dfrac{6x}{x + 1}$
$\ln \dfrac{x^2}{3(x + 1)}$
$\dfrac{\ln 3x^2}{\ln(x + 1)}$

Power rule first: $2\ln x = \ln x^2$.

Added logs multiply: $\ln x^2 + \ln 3 = \ln 3x^2$. A subtracted log divides: $\ln 3x^2 - \ln(x + 1) = \ln \dfrac{3x^2}{x + 1}$.

Answer: $\ln \dfrac{3x^2}{x + 1}$.

Solve $9^{x + 1} = 27^x$.

Write both sides with base 3: $9 = 3^2$ and $27 = 3^3$.

$(3^2)^{x + 1} = (3^3)^x$, so $3^{2x + 2} = 3^{3x}$.

Same base, so the exponents are equal: $2x + 2 = 3x$, which gives $x = 2$.

Check: $9^3 = 729$ and $27^2 = 729$. Answer: $x = 2$.

An account grows by $A = 1000e^{0.06t}$, where $t$ is in years. How many years until it doubles to $2,000? Use a calculator and round to the nearest tenth of a year.

Set $1000e^{0.06t} = 2000$ and divide by 1000: $e^{0.06t} = 2$.

Take $\ln$ of both sides: $0.06t = \ln 2$.

$t = \dfrac{\ln 2}{0.06} \approx \dfrac{0.6931}{0.06} \approx 11.55$. Answer: about $11.6$ years.

Solve $\log_5(x + 3) = 2$.

Rewrite in exponential form: $5^2 = x + 3$.

$25 = x + 3$, so $x = 22$.

Check: $\log_5 25 = 2$. Answer: $x = 22$.

Solve $\log_2 x + \log_2(x - 2) = 3$. Check for extraneous solutions.

Combine with the product rule: $\log_2\big(x(x - 2)\big) = 3$.

Exponential form: $x(x - 2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$, which factors as $(x - 4)(x + 2) = 0$.

$x = 4$ or $x = -2$. But $\log_2(-2)$ is undefined, so $-2$ is extraneous.

Check $x = 4$: $\log_2 4 + \log_2 2 = 2 + 1 = 3$. Answer: $x = 4$ only.

Multiply and simplify: $\dfrac{x^2 - 9}{2x + 4} \cdot \dfrac{x + 2}{x^2 - 3x}$.

Factor everything: $\dfrac{(x - 3)(x + 3)}{2(x + 2)} \cdot \dfrac{x + 2}{x(x - 3)}$.

Cancel the common factors $(x - 3)$ and $(x + 2)$.

What is left: $\dfrac{x + 3}{2x}$. Answer: $\dfrac{x + 3}{2x}$.

Factor: $x^3 - 27$.

It is a difference of cubes: $x^3 - 3^3$, with $a = x$ and $b = 3$.

Pattern: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

$(x - 3)(x^2 + 3x + 9)$. Answer: $(x - 3)(x^2 + 3x + 9)$.

Want to look something up first? Exponentials and logs cheat sheet · Glossary

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