Exponential and Logarithmic Function Practice Questions

The expression \(\log_2 32\) asks: “2 raised to what power equals 32?” Rewrite it as an exponential equation:

\(2^x = 32\)

Since \(2^5 = 32\), the answer is 5.

In general, \(\log_b a = c\) means \(b^c = a\). The logarithm is the exponent.

The exponential form \(b^c = a\) converts to logarithmic form \(\log_b a = c\). The base stays the base, the result goes inside the log, and the exponent becomes the answer:

\(5^3 = 125 \implies \log_5 125 = 3\)

Choice B puts the exponent (3) as the base of the logarithm. Choice C swaps the base and the result. Remember: the base in the exponential equation is always the base in the logarithm.

Use the product rule: \(\log_b m + \log_b n = \log_b(mn)\):

\(\log_3 9 + \log_3 27 = \log_3(9 \times 27) = \log_3 243\)

Since \(3^5 = 243\), the answer is 5.

Alternatively, evaluate each logarithm separately: \(\log_3 9 = 2\) (since \(3^2 = 9\)) and \(\log_3 27 = 3\) (since \(3^3 = 27\)). So \(2 + 3 = 5\).

Apply logarithm properties step by step:

Step 1: Quotient Rule

\(\log_b \frac{m}{n} = \log_b m – \log_b n\)

\(\log_2 \dfrac{x^3}{y} = \log_2 x^3 – \log_2 y\)

Step 2: Power Rule

\(\log_b m^n = n\log_b m\)

\(= 3\log_2 x – \log_2 y\)

Choice B writes \(\log_2 3x\) instead of \(3\log_2 x\). The power comes in front as a coefficient, not multiplied inside the argument. Choice C incorrectly applies the 3 to both terms.

Apply the properties in reverse order of expanding:

Step 1: Power Rule

Move coefficients up as exponents:

\(\log x^2 + \log y – \log z^3\)

Step 2: Product Rule

Addition becomes multiplication:

\(\log(x^2 y) – \log z^3\)

Step 3: Quotient Rule

Subtraction becomes division:

\(= \log \dfrac{x^2 y}{z^3}\)

Choice B treats the coefficients as multipliers inside the log instead of exponents. Choice C uses addition inside the argument instead of multiplication.

Rewrite both sides with the same base. Since \(16 = 2^4\):

\(2^{x+1} = 2^4\)

When the bases are equal, the exponents must be equal:

\(x + 1 = 4 \implies x = 3\)

Verify: \(2^{3+1} = 2^4 = 16\) ✓

Convert the logarithmic equation to exponential form. Remember, \(\log_b a = c\) means \(b^c = a\):

\(\log_4 x = 3 \implies 4^3 = x \implies x = 64\)

Choice A multiplies the base by the exponent (\(4 \times 3 = 12\)) instead of raising to the power. Choice D computes \(4^4 = 256\) instead of \(4^3\).

Since 40 is not a power of 5, take the logarithm of both sides. You can use either common log or natural log (the result is the same). Using common log:

\(\log(5^x) = \log 40\)

Apply the power rule:

\(x \cdot \log 5 = \log 40\)

Solve for \(x\):

\(x = \dfrac{\log 40}{\log 5} \approx \dfrac{1.602}{0.699} \approx 2.292\)

This is also equivalent to using the change of base formula: \(x = \log_5 40 = \frac{\log 40}{\log 5}\).

Choice B gives the value of \(\log 40\) alone, without dividing by \(\log 5\).

The natural logarithm \(\ln\) has base \(e\). A fundamental property of logarithms is that \(\log_b b^x = x\). The logarithm and exponential are inverse operations that cancel each other:

\(\ln e^7 = \log_e e^7 = 7\)

Similarly, \(e^{\ln x} = x\). These inverse properties are essential for solving exponential and logarithmic equations.

The argument of a logarithm must be strictly positive (greater than zero, not equal to zero). Set the argument greater than zero and solve:

\(x – 4 \gt 0 \implies x \gt 4\)

Choice B uses \(\geq\) instead of \(\gt\). However, \(\log_3(0)\) is undefined, so \(x = 4\) must be excluded. This is different from square roots, where the argument can equal zero.

Use the product rule to combine the left side:

\(\log[x(x – 3)] = 1\)

Convert to exponential form (base 10, since \(\log\) means \(\log_{10}\)):

\(x(x – 3) = 10^1 = 10\)

Expand and solve the quadratic:

\(x^2 – 3x – 10 = 0\)

\((x – 5)(x + 2) = 0\)

This gives \(x = 5\) or \(x = -2\). However, check both in the original equation:

• \(x = 5\): \(\log 5 + \log 2 = \log 10 = 1\) ✓
• \(x = -2\): \(\log(-2)\) is undefined ✗

Since \(x = -2\) produces a negative argument inside the logarithm, it is extraneous. The only valid solution is \(x = 5\).

The general form for exponential growth is \(P(t) = P_0 \cdot b^{t/k}\), where \(P_0\) is the initial amount, \(b\) is the growth factor, and \(k\) is the time it takes for one cycle of growth.

Here, \(P_0 = 500\), the growth factor is 2 (doubling), and it takes 3 hours per doubling:

\(P(t) = 500 \cdot 2^{t/3}\)

Choice A omits the division by 3, which would mean the population doubles every hour instead of every 3 hours. Choice D models linear growth, not exponential.

Depreciation is modeled by exponential decay. Each year the car retains \(100\% – 15\% = 85\%\) of its value, so the decay factor is 0.85:

\(V(t) = 25{,}000 \cdot (0.85)^t\)

Substitute \(t = 4\):

\(V(4) = 25{,}000 \cdot (0.85)^4\)\(\:= 25{,}000 \cdot 0.52200625\)\(\:\approx \$13{,}050.16\)

Choice D computes only one year of depreciation (\(25{,}000 \times 0.85 = 21{,}250\)). Choice C subtracts 15% of the original value four times (linear depreciation), which is not how exponential decay works.

Logarithmic and exponential functions are inverses of each other. To find the inverse, swap \(x\) and \(y\) and solve:

\(y = 3^x \implies x = 3^y \implies y = \log_3 x\)

So \(f^{-1}(x) = \log_3 x\). You can verify:

\(f(f^{-1}(x)) = 3^{\log_3 x} = x\)

Choice C gives the reciprocal of \(f(x)\), not the inverse. Choice D gives a reflection over the \(y\)-axis, not the inverse. Remember: inverse function ≠ reciprocal.

The change of base formula converts a logarithm of any base into a ratio of logarithms with a different base:

\(\log_b a = \dfrac{\log a}{\log b}\)

Apply this to \(\log_5 18\):

\(\log_5 18 = \dfrac{\log 18}{\log 5}\)

This works equally well with natural logarithms: \(\log_5 18 = \frac{\ln 18}{\ln 5}\). The change of base formula is especially useful when evaluating logarithms on a calculator, which typically only has \(\log\) (base 10) and \(\ln\) (base \(e\)) buttons.

Choice B flips the fraction. Choice D uses subtraction (\(\log \frac{18}{5}\)) instead of division, confusing the quotient rule with the change of base formula.