The properties of logarithms allow us to rewrite complex expressions in a simpler form by means of the product, quotient and power rules. This collection presents 20 exercises of progressive difficulty, each with a fully worked-out, commented solution.
Exercise 1 — level ★☆☆☆☆
\[ \log_2(4 \cdot 8) \]
Answer
\[ 5 \]
Solution
Apply the product rule for logarithms:
\[ \log_2(4 \cdot 8) = \log_2 4 + \log_2 8 \]
Evaluate each logarithm separately: \[ \log_2 4 = 2, \quad \log_2 8 = 3 \]
Add the results: \[ 2 + 3 = 5 \]
Exercise 2 — level ★☆☆☆☆
\[ \log_3\left(\frac{81}{3}\right) \]
Answer
\[ 3 \]
Solution
Use the quotient rule:
\[ \log_3\left(\frac{81}{3}\right) = \log_3 81 - \log_3 3 \]
Evaluate the logarithms: \[ \log_3 81 = 4, \quad \log_3 3 = 1 \]
Subtract: \[ 4 - 1 = 3 \]
Exercise 3 — level ★☆☆☆☆
\[ \log_5(25^3) \]
Answer
\[ 6 \]
Solution
Apply the power rule:
\[ \log_5(25^3) = 3\log_5 25 \]
Since \(25 = 5^2\), we have:
\[ \log_5 25 = 2 \]
Therefore:
\[ 3 \cdot 2 = 6 \]
Exercise 4 — level ★★☆☆☆
\[ \log_2\sqrt{32} \]
Answer
\[ \frac{5}{2} \]
Solution
Rewrite the radical as a power:
\[ \sqrt{32} = 32^{1/2} \]
Factor 32:
\[ 32 = 2^5 \]
Hence:
\[ (2^5)^{1/2} = 2^{5/2} \]
Apply the logarithm:
\[ \log_2(2^{5/2}) = \frac{5}{2} \]
Exercise 5 — level ★★☆☆☆
\[ \log_3\left(\frac{1}{27}\right) \]
Answer
\[ -3 \]
Solution
Express 27 as a power of 3:
\[ 27 = 3^3 \Rightarrow \frac{1}{27} = 3^{-3} \]
Apply the logarithm:
\[ \log_3(3^{-3}) = -3 \]
Exercise 6 — level ★★☆☆☆
\[ \ln(e^2 \cdot \sqrt{e}) \]
Answer
\[ \frac{5}{2} \]
Solution
Rewrite the radical as a power:
\[ \sqrt{e} = e^{1/2} \]
Apply the product rule for exponents:
\[ e^2 \cdot e^{1/2} = e^{5/2} \]
Finally, apply the natural logarithm:
\[ \ln(e^{5/2}) = \frac{5}{2} \]
Exercise 7 — level ★★☆☆☆
\[ \log(100x) \]
Answer
\[ 2 + \log x \]
Solution
Apply the product rule for logarithms:
\[ \log(100x) = \log 100 + \log x \]
Evaluate the numerical logarithm:
\[ \log 100 = 2 \]
Substituting back, we obtain:
\[ 2 + \log x \]
Exercise 8 — level ★★★☆☆
\[ 2\log a + 3\log b \]
Answer
\[ \log(a^2 b^3) \]
Solution
Use the power rule for logarithms:
\[ 2\log a = \log(a^2), \quad 3\log b = \log(b^3) \]
Rewrite the expression:
\[ \log(a^2) + \log(b^3) \]
Apply the product rule:
\[ \log(a^2 b^3) \]
Exercise 9 — level ★★★☆☆
\[ \log_b\left(\frac{x^2}{y}\right) \]
Answer
\[ 2\log_b x - \log_b y \]
Solution
Apply the quotient rule:
\[ \log_b\left(\frac{x^2}{y}\right) = \log_b(x^2) - \log_b(y) \]
Use the power rule:
\[ \log_b(x^2) = 2\log_b x \]
Substituting:
\[ 2\log_b x - \log_b y \]
Exercise 10 — level ★★★☆☆
\[ \log_4 8 \]
Answer
\[ \frac{3}{2} \]
Solution
Apply the change-of-base formula:
\[ \log_4 8 = \frac{\log_2 8}{\log_2 4} \]
Evaluate the logarithms:
\[ \log_2 8 = 3, \quad \log_2 4 = 2 \]
Divide:
\[ \frac{3}{2} \]
Exercise 11 — level ★★★☆☆
\[ \log_2 6 + \log_2 4 - \log_2 3 \]
Answer
\[ 3 \]
Solution
Apply the sum and difference rules:
\[ \log_2 6 + \log_2 4 = \log_2(24) \]
Subtract the third logarithm:
\[ \log_2\left(\frac{24}{3}\right) \]
Simplify:
\[ \log_2 8 = 3 \]
Exercise 12 — level ★★★★☆
\[ \log_b \sqrt[3]{\frac{a}{b}} \]
Answer
\[ \frac{1}{3}\log_b a - \frac{1}{3} \]
Solution
Rewrite the radical as a power:
\[ \sqrt[3]{\frac{a}{b}} = \left(\frac{a}{b}\right)^{1/3} \]
Apply the power rule:
\[ \log_b\left(\frac{a}{b}\right)^{1/3} = \frac{1}{3}\log_b\left(\frac{a}{b}\right) \]
Use the quotient rule:
\[ \log_b a - \log_b b \]
Since \(\log_b b = 1\), substitute:
\[ \frac{1}{3}(\log_b a - 1) \]
Exercise 13 — level ★★★★☆
\[ \frac{1}{2}\log x - 2\log y - 3\log z \]
Answer
\[ \log\left(\frac{\sqrt{x}}{y^2 z^3}\right) \]
Solution
Apply the power rule:
\[ \frac{1}{2}\log x = \log(x^{1/2}), \quad 2\log y = \log(y^2), \quad 3\log z = \log(z^3) \]
Rewrite the expression:
\[ \log(x^{1/2}) - \log(y^2) - \log(z^3) \]
Combine using the properties of logarithms:
\[ \log\left(\frac{x^{1/2}}{y^2 z^3}\right) \]
Exercise 14 — level ★★★★☆
\[ \log_2(x^2 - 1) - \log_2(x - 1) \]
Answer
\[ \log_2(x+1), \quad x>1 \]
Solution
Apply the quotient rule:
\[ \log_2\left(\frac{x^2 - 1}{x - 1}\right) \]
Factor the difference of squares:
\[ x^2 - 1 = (x-1)(x+1) \]
Cancel and simplify:
\[ \log_2(x+1) \]
Exercise 15 — level ★★★★☆
\[ \log_{1/2} 16 \]
Answer
\[ -4 \]
Solution
Apply the change of base:
\[ \log_{1/2} 16 = \frac{\log_2 16}{\log_2(1/2)} \]
Evaluate the values:
\[ \log_2 16 = 4, \quad \log_2(1/2) = -1 \]
Divide:
\[ -4 \]
Exercise 16 — level ★★★★★
\[ e^{-2\ln x} \]
Answer
\[ \frac{1}{x^2} \]
Solution
Use the property:
\[ -2\ln x = \ln(x^{-2}) \]
Rewrite the expression:
\[ e^{\ln(x^{-2})} \]
Simplify:
\[ x^{-2} = \frac{1}{x^2} \]
Exercise 17 — level ★★★★★
\[ \log \sqrt{x\sqrt{x}} \]
Answer
\[ \frac{3}{4}\log x \]
Solution
Rewrite the inner radical as a power:
\[ \sqrt{x} = x^{1/2} \]
The expression becomes:
\[ \sqrt{x \cdot x^{1/2}} \]
Add the exponents:
\[ x \cdot x^{1/2} = x^{3/2} \]
Apply the outer radical:
\[ (x^{3/2})^{1/2} = x^{3/4} \]
Apply the logarithm:
\[ \log(x^{3/4}) = \frac{3}{4}\log x \]
Exercise 18 — level ★★★★★
\[ (\log_3 5)(\log_5 9) \]
Answer
\[ 2 \]
Solution
Apply the change-of-base formula:
\[ \log_3 5 = \frac{\ln 5}{\ln 3}, \quad \log_5 9 = \frac{\ln 9}{\ln 5} \]
Multiply the expressions:
\[ \frac{\ln 5}{\ln 3} \cdot \frac{\ln 9}{\ln 5} \]
Cancel the common factor \(\ln 5\):
\[ \frac{\ln 9}{\ln 3} \]
Since \(9 = 3^2\), we obtain:
\[ \log_3 9 = 2 \]
Exercise 19 — level ★★★★★
\[ \ln\left(\frac{e^x}{1+e^x}\right) \]
Answer
\[ x - \ln(1+e^x) \]
Solution
Apply the quotient rule for logarithms:
\[ \ln\left(\frac{e^x}{1+e^x}\right) = \ln(e^x) - \ln(1+e^x) \]
Simplify the first term:
\[ \ln(e^x) = x \]
Hence we obtain:
\[ x - \ln(1+e^x) \]
Exercise 20 — level ★★★★★
\[ \log_b\left(\frac{1}{\sqrt[n]{b^m}}\right) \]
Answer
\[ -\frac{m}{n} \]
Solution
Rewrite the radical as a power:
\[ \sqrt[n]{b^m} = b^{m/n} \]
Therefore:
\[ \frac{1}{\sqrt[n]{b^m}} = b^{-m/n} \]
Apply the logarithm:
\[ \log_b(b^{-m/n}) = -\frac{m}{n} \]