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Exponentials & logarithms: Algebra 2 Regents Practice

Practice Exponentials & logarithms for the Algebra 2 Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 33 · 2 pts

Evaluate $\log_3 81$.

Worked solution
$\log_3 81 = x \iff 3^x = 81$ Rewrite in exponential form.
$81 = 3^4$
$3^x = 3^4 \implies x = 4$
Answer: $4$

A logarithm asks 'what exponent gives this value?'; since $3^4=81$, $\log_3 81 = 4$.

Question 34 · 2 pts

Solve for $x$: $2^{x+1} = 32$.

Worked solution
$32 = 2^5$ Write both sides with base $2$.
$2^{x+1} = 2^5$
$x + 1 = 5$ Equal bases means equal exponents.
$x = 4$
Answer: $x = 4$

With equal bases, set the exponents equal; $2^5=32$ gives $x+1=5$, so $x=4$.

Question 44 · 2 pts

A radioactive sample has a half-life of 5 years, modeled by $A(t) = 80\left(\tfrac{1}{2}\right)^{t/5}$, where $t$ is in years. How many grams remain after 15 years?

Worked solution
$A(15) = 80\left(\tfrac{1}{2}\right)^{15/5}$
$= 80\left(\tfrac{1}{2}\right)^{3}$ 15 years is exactly 3 half-lives.
$= 80 \cdot \tfrac{1}{8} = 10$
Answer: $10$ grams

In a half-life model the exponent $t/5$ counts how many half-lives have elapsed; here 15 years equals 3 half-lives, so the amount is halved three times.

Question 48 · 2 pts

Find the average rate of change of $f(x) = 2^{x}$ over the interval $[1, 3]$.

Worked solution
$f(1) = 2^{1} = 2, \quad f(3) = 2^{3} = 8$ Evaluate the function at each endpoint.
$\text{AROC} = \dfrac{f(3) - f(1)}{3 - 1}$
$= \dfrac{8 - 2}{2} = \dfrac{6}{2} = 3$
Answer: $3$

The average rate of change over an interval is the slope of the secant line: the change in the function values divided by the change in $x$.

Question 4 · 2 pts

The value of $16^{\frac{3}{4}}$ is

Worked solution
$16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}}$ Write 16 as a power of 2.
$(2^4)^{\frac{3}{4}} = 2^{4 \cdot \frac{3}{4}} = 2^3$ Multiply the exponents.
$2^3 = 8$
Answer: $8$

A rational exponent means a root and a power: $16^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8$. Multiplying $16$ by $\frac{3}{4}$ to get 12 is the most common error.

Question 2 · 2 pts

For all positive values of $x$, the expression $x^{\frac{2}{3}}$ is equivalent to

Worked solution
$x^{\frac{m}{n}}=\sqrt[n]{x^m}$ The denominator of a rational exponent is the index of the root.
$x^{\frac{2}{3}}=\sqrt[3]{x^2}$
Answer: $\sqrt[3]{x^2}$

A rational exponent $\frac{m}{n}$ means the $n$th root of the $m$th power. Here the denominator 3 is the index of the root and the numerator 2 is the power, so $x^{2/3}=\sqrt[3]{x^2}$. Swapping the roles of 2 and 3 gives the common wrong answer $\sqrt{x^3}$.

Question 14 · 2 pts

Which expression is equivalent to $(27x^6)^{\frac{2}{3}}$?

Worked solution
$(27x^6)^{\frac{2}{3}} = 27^{\frac{2}{3}} \cdot (x^6)^{\frac{2}{3}}$ Apply the power to each factor.
$27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$ The denominator of the exponent is the root.
$(x^6)^{\frac{2}{3}} = x^{6 \cdot \frac{2}{3}} = x^4$ Multiply exponents.
$(27x^6)^{\frac{2}{3}} = 9x^4$
Answer: $9x^4$

A rational exponent of $\frac{2}{3}$ means take the cube root, then square. The cube root of 27 is 3, which squared is 9, and the exponent on $x$ is $6 \cdot \frac{2}{3} = 4$, giving $9x^4$.

Question 26 · 4 pts · constructed response

Solve algebraically for $x$: $8^{2x} = 4^{x+3}$

Model answer: $x = \frac{3}{2}$
Worked solution
$8^{2x} = (2^3)^{2x} = 2^{6x}$ Rewrite 8 as a power of 2.
$4^{x+3} = (2^2)^{x+3} = 2^{2x+6}$ Rewrite 4 as a power of 2.
$2^{6x} = 2^{2x+6} \Rightarrow 6x = 2x + 6$ Equal bases mean the exponents are equal.
$4x = 6 \Rightarrow x = \frac{3}{2}$
Answer: $x = \frac{3}{2}$

Both sides can be written with the common base 2. Once the bases match, the exponential function is one-to-one, so the exponents must be equal, giving a simple linear equation with solution $x = \frac{3}{2}$.

Question 34 · 4 pts · constructed response

Solve the equation $4^{2x+1} = 9$ for $x$ algebraically. Express your answer to the nearest hundredth.

Model answer: $x = \dfrac{1}{2}\left(\dfrac{\ln 9}{\ln 4} - 1\right) \approx 0.29$
Worked solution
$4^{2x+1} = 9$ Take the natural log (or common log) of both sides.
$(2x+1)\ln 4 = \ln 9$ Use the power rule for logarithms.
$2x + 1 = \frac{\ln 9}{\ln 4} \approx 1.58496$
$2x \approx 0.58496$
$x \approx 0.29$ Rounded to the nearest hundredth.
Answer: $x = \dfrac{1}{2}\left(\dfrac{\ln 9}{\ln 4} - 1\right) \approx 0.29$

Because the variable is in the exponent, taking the logarithm of both sides lets the power rule bring the exponent down. Solving the resulting linear equation gives $x = \frac{1}{2}(\log_4 9 - 1) \approx 0.29$. Keeping extra decimal places until the final step avoids rounding error.

Question 15 · 2 pts

What is the solution to the equation $\log_3(x-2)=4$?

Worked solution
$\log_3(x-2)=4 \;\Rightarrow\; x-2=3^4$ Rewrite the log equation in exponential form.
$x-2=81$
$x=83$ Add 2 to both sides.
Answer: $83$

A logarithmic equation is solved by converting to exponential form: $\log_b a = c$ means $b^c = a$. Here $x-2 = 3^4 = 81$, so $x = 83$. Forgetting to add back the 2 gives the distractor 81.

Question 19 · 2 pts

What is the value of $\log_2 32$ ?

Worked solution
$\log_2 32 = x \Rightarrow 2^x = 32$ A log is the exponent.
$2^5 = 32 \Rightarrow x = 5$
Answer: 5

log_b(n) asks: to what power must b be raised to get n? Here 2⁵ = 32.

Question 35 · 2 pts

Solve for $x$: $\log_2(x-3) = 4$.

Worked solution
$\log_2(x-3) = 4 \iff x - 3 = 2^4$ Convert to exponential form.
$x - 3 = 16$
$x = 19$
Answer: $x = 19$

Rewriting the log as $x-3=2^4$ undoes it; $16+3=19$.

Question 37 · 2 pts

Solve $3^x = 20$ for $x$, rounded to the nearest hundredth.

Worked solution
$3^x = 20 \iff x = \log_3 20$ Rewrite in log form.
$x = \frac{\log 20}{\log 3}$ Change-of-base formula.
$x = \frac{1.3010}{0.4771} \approx 2.73$
Answer: $x \approx 2.73$

Take a log of both sides and apply change of base: $x=\frac{\log 20}{\log 3}\approx 2.73$.

Question 38 · 2 pts

The value of a car is modeled by $V(t) = 24000(0.88)^t$, where $t$ is in years. What is the annual rate of depreciation?

Worked solution
$V(t) = 24000(0.88)^t$ Compare to $A(1+r)^t$.
$1 + r = 0.88$
$r = -0.12$ Negative means decay.
$|r| = 0.12 = 12\%$
Answer: $12\%$ per year

In $A(1+r)^t$, a base of $0.88$ means $1+r=0.88$, so $r=-0.12$: a $12\%$ annual decrease.

Question 40 · 2 pts

Write $\log_b\!\left(\dfrac{x^2\sqrt{y}}{z^3}\right)$ as a sum and difference of logarithms.

Worked solution
$\log_b(x^2\sqrt{y}) - \log_b(z^3)$ Quotient rule.
$\log_b x^2 + \log_b y^{1/2} - \log_b z^3$ Product rule; $\sqrt{y}=y^{1/2}$.
$2\log_b x + \tfrac{1}{2}\log_b y - 3\log_b z$ Power rule.
Answer: $2\log_b x + \frac{1}{2}\log_b y - 3\log_b z$

Apply the quotient, product, and power rules: exponents become coefficients and division becomes subtraction.

Question 41 · 2 pts

Solve for $x$: $\;9^{\,2x-1} = 27^{\,x+2}$.

Worked solution
$9 = 3^{2} \quad\text{and}\quad 27 = 3^{3}$ Rewrite both sides with the common base 3.
$\left(3^{2}\right)^{2x-1} = \left(3^{3}\right)^{x+2}$
$3^{\,4x-2} = 3^{\,3x+6}$ Multiply the exponents.
$4x - 2 = 3x + 6$ Equal bases force equal exponents.
$x = 8$
Answer: $x = 8$

When an exponential equation has different bases that are powers of the same number, rewrite each side with that common base, then set the exponents equal.

Question 42 · 2 pts

Express $\;2\log x + \log 3 - \log y\;$ as a single logarithm.

Worked solution
$2\log x = \log x^{2}$ Power rule: a coefficient becomes an exponent.
$\log x^{2} + \log 3 = \log\left(3x^{2}\right)$ Product rule: sum of logs is log of the product.
$\log\left(3x^{2}\right) - \log y = \log\dfrac{3x^{2}}{y}$ Quotient rule: difference of logs is log of the quotient.
Answer: $\log\dfrac{3x^{2}}{y}$

Condensing logs uses three properties: the power rule turns coefficients into exponents, addition becomes multiplication, and subtraction becomes division.

Question 46 · 2 pts

What is the domain of $f(x) = \log_{5}(2x - 6)$?

Worked solution
$2x - 6 > 0$ The argument of a logarithm must be positive.
$2x > 6$
$x > 3$
Answer: $x > 3$

A logarithm is defined only when its argument is positive, so the domain is found by solving the inequality argument $> 0$.

Question 47 · 2 pts

Which expression gives the exact solution of $\;7^{\,x} = 20$?

Worked solution
$7^{x} = 20$
$x = \log_{7} 20$ Definition of a logarithm: b^x = a means x = log_b(a).
$x = \dfrac{\ln 20}{\ln 7} \approx 1.54$ Change-of-base gives a decimal if needed.
Answer: $x = \log_{7} 20$

When an exponential equation cannot be written with a common base, take a logarithm; by definition the solution of $b^{x}=a$ is $x = \log_{b} a$.

Question 3 · 2 pts

Which expression is equivalent to $\log_3 x + 2\log_3 y - \log_3 z$?

Worked solution
$2\log_3 y = \log_3 y^2$ Power rule: a coefficient becomes an exponent.
$\log_3 x + \log_3 y^2 = \log_3\left(xy^2\right)$ Sum of logs is the log of a product.
$\log_3\left(xy^2\right) - \log_3 z = \log_3\dfrac{xy^2}{z}$ Difference of logs is the log of a quotient.
Answer: $\log_3\dfrac{xy^2}{z}$

Apply the power rule first so the coefficient 2 becomes the exponent on $y$ only, then combine with the product and quotient rules. Distractors come from multiplying by 2 instead of raising to a power, or applying the exponent to both $x$ and $y$.

Question 7 · 2 pts

For $x>0$, which expression is equivalent to $\sqrt[3]{x^2}\cdot\sqrt{x}$?

Worked solution
$\sqrt[3]{x^2} = x^{2/3},\qquad \sqrt{x} = x^{1/2}$ Rewrite each radical with a rational exponent.
$x^{2/3}\cdot x^{1/2} = x^{2/3+1/2}$ Add exponents when multiplying like bases.
$\dfrac{2}{3}+\dfrac{1}{2} = \dfrac{4}{6}+\dfrac{3}{6} = \dfrac{7}{6}$
$x^{7/6} = \sqrt[6]{x^7}$
Answer: $\sqrt[6]{x^7}$

Convert each radical to a rational exponent and add the exponents: $\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$, so the product is $x^{7/6}=\sqrt[6]{x^7}$. Multiplying the exponents gives $\frac{1}{3}$, and adding numerators and denominators separately gives $\frac{3}{5}$, producing the distractors.

Question 15 · 2 pts

What is the solution of the equation $4^x = 8^{x-1}$?

Worked solution
$4^x = (2^2)^x = 2^{2x}$ Write both sides with base 2
$8^{x-1} = (2^3)^{x-1} = 2^{3x-3}$
$2x = 3x - 3$ Equal bases mean equal exponents
$x = 3$
$\text{Check: } 4^3 = 64 \text{ and } 8^{2} = 64$
Answer: $x = 3$

Rewriting both sides as powers of 2 gives $2^{2x} = 2^{3x-3}$. With a common base, the exponents must be equal, so $2x = 3x - 3$, which gives $x = 3$. Both sides then equal 64.

Question 32 · 4 pts · constructed response

Solve algebraically for $x$: $\log_3(x+5) + \log_3(x-1) = 3$

Model answer: $x = 4$ (the value $x=-8$ is rejected because it makes the arguments of the logarithms negative)
Worked solution
$\log_3\big[(x+5)(x-1)\big] = 3$ Use the product rule for logarithms.
$(x+5)(x-1) = 3^3 = 27$ Rewrite in exponential form.
$x^2 + 4x - 5 = 27 \Rightarrow x^2 + 4x - 32 = 0$
$(x+8)(x-4) = 0 \Rightarrow x = -8 \ \text{or} \ x = 4$ Factor and solve.
$x = -8 \ \text{is rejected since} \ x - 1 < 0; \ \text{so} \ x = 4$ Check the domain of the original logarithms.
Answer: $x = 4$ (the value $x=-8$ is rejected because it makes the arguments of the logarithms negative)

Combining the two logarithms with the product rule and converting to exponential form gives a quadratic equation. Both roots must be checked in the original equation: $x=-8$ makes the log arguments negative and is extraneous, so the only solution is $x=4$.

Question 35 · 4 pts · constructed response

The number of bacteria in a culture is modeled by $P(t) = 500(1.03)^t$, where $t$ is the time in hours since the culture was started. Determine algebraically, to the nearest tenth of an hour, how long it will take for the population to double.

Model answer: About $23.4$ hours, since $t = \dfrac{\log 2}{\log 1.03} \approx 23.4$.
Worked solution
$1000 = 500(1.03)^t$ Doubling means the population reaches 2(500) = 1000.
$2 = (1.03)^t$ Divide both sides by 500.
$\log 2 = t \log 1.03$ Take the logarithm of both sides and apply the power rule.
$t = \frac{\log 2}{\log 1.03} \approx 23.4498$ Any base works; natural log gives the same value.
$t \approx 23.4 \text{ hours}$
Answer: About $23.4$ hours, since $t = \dfrac{\log 2}{\log 1.03} \approx 23.4$.

Doubling time does not depend on the starting amount: dividing by 500 leaves $2 = 1.03^t$. Since the variable is in the exponent, logarithms are needed to solve, and the power rule brings $t$ down, giving $t = \log 2 / \log 1.03 \approx 23.4$ hours.

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