LOGARITHM REVISION QUESTIONS WITH ANSWERS

Logarithm Questions

High School Logarithm Questions

Evaluate \( \log_{10} 1000 \)
Simplify: \( \log 2 + \log 5 \)
If \( \log x = 2 \), find the value of \( x \)
Solve: \( \log_{2} x = 5 \)
Using tables, find \( \log_{10} 32.5 \)
Find \( \log_{10} 0.00345 \) using tables
Prove: \( \log(ab) = \log a + \log b \)
Simplify: \( \log 50 - \log 2 \)
Evaluate \( \log_{10}(100 \times 0.01) \)
Express \( \log 8 \) in terms of \( \log 2 \)
If \( \log x = a \), express \( x \) in terms of \( a \)
Simplify: \( 2\log 3 - \log 9 \)
Evaluate: \( \log_{10} \left(\frac{25}{4}\right) \)
Solve: \( \log(x^2) = 6 \)
If \( \log 2 = 0.3010 \), find \( \log 16 \)
Using log tables, find the value of \( 73.4 \times 0.052 \)
Find the number of digits in \( 2^{10} \)
Solve for \( x \): \( \log_{3}(x - 2) = 2 \)
Simplify: \( \log_5(125) \)
Evaluate: \( \log(10^{-3}) \)
If \( \log a = x \) and \( \log b = y \), express \( \log \left(\frac{a^2}{\sqrt{b}}\right) \) in terms of \( x \) and \( y \)
Find the antilog of 2.643
Using tables, calculate \( \frac{465 \times 0.0732}{9.52} \)
Evaluate: \( \log_{4} 64 \)
Solve for \( x \): \( \log_{x} 16 = 4 \)
Simplify: \( \log(1000) - \log(10) \)
If \( \log 7 = 0.8451 \), find \( \log 49 \)
Prove: \( \log_b a = \frac{\log a}{\log b} \)
Solve: \( \log_{2}(x+3) = \log_{2}(x-1) + 1 \)
Find \( x \) if \( \log_{10} x = -2.3 \)

3
1
100
32
1.5119
−2.461
By law: \( \log(ab) = \log a + \log b \)
\( \log 25 = 1.6989 \)
0
\( \log 8 = 3\log 2 \)
\( x = 10^a \)
0
\( \log 25 - \log 4 = \log \frac{25}{4} \approx 0.3979 \)
\( x = 10^3 = 1000 \)
1.2040
\( \log(73.4) + \log(0.052) = 1.865 + (-1.284) = 0.581 \Rightarrow \text{antilog } 0.581 \approx 3.81 \)
4 digits (since \( \log(2^{10}) = \log(1024) \approx 3.01 \))
11
3
−3
\( 2x - \frac{1}{2}y \)
\( \text{antilog}(2.643) \approx 441.7 \)
\( \log(465) + \log(0.0732) - \log(9.52) = 2.668 + (-1.135) - 0.979 \approx 0.554 \Rightarrow \text{antilog} \approx 3.58 \)
3
2
2
\( 2\log 7 = 2 \times 0.8451 = 1.6902 \)
\( \log_b a = \frac{\log a}{\log b} \) — Change of base law
\( x = 5 \)
\( x = 10^{-2.3} \approx 0.00501 \)