Simplify \( \dfrac{a^5 \times a^{-2}}{a^3} \).
\( a^{5-2-3} = a^0 = 1 \).
Exponentials & Logarithms
12 exam-style questions on laws of indices and logs, equations and modelling. Tap Show answer for full working.
4 Easy4 Medium4 Hard
Evaluate \( \log_{10} 1000 \).
\( 3 \).
Solve \( 2^x = 32 \).
\( x = 5 \).
Write \( \log 12 - \log 3 \) as a single logarithm.
\( \log 4 \).
Solve \( 3^{2x} = 27 \).
\( 3^{2x} = 3^3 \) ⇒ \( 2x=3 \) ⇒ \( x=3/2 \).
Solve \( \log_2 (x+1) = 3 \).
\( x+1 = 2^3 = 8 \) ⇒ \( x=7 \).
Express \( 2\log a + 3\log b - \log c \) as a single logarithm.
\( \log \dfrac{a^2 b^3}{c} \).
Solve \( e^{2x} - 3e^x + 2 = 0 \).
Let \( u=e^x \): \( u^2 - 3u + 2=0 \) ⇒ \( (u-1)(u-2)=0 \) ⇒ \( e^x=1 \) or \( 2 \) ⇒ \( x=0 \) or \( x=\ln 2 \).
Solve \( \log_3 (x-1) + \log_3 (x+1) = 1 \).
\( \log_3 ((x-1)(x+1)) = 1 \) ⇒ \( x^2-1 = 3 \) ⇒ \( x^2=4 \) ⇒ \( x=2 \) (since \( x>1 \)).
The population of a colony of bacteria is modelled by \( P = 500 e^{0.2t} \) where \( t \) is in hours. Find the time taken for the population to double.
\( 1000 = 500 e^{0.2t} \) ⇒ \( 2 = e^{0.2t} \) ⇒ \( \ln 2 = 0.2t \) ⇒ \( t = 5\ln 2 \) hours.
Solve \( 2^{x+1} = 5^{x-1} \), giving your answer in terms of logarithms.
Take ln: \( (x+1)\ln 2 = (x-1)\ln 5 \) ⇒ \( x\ln 2 + \ln 2 = x\ln 5 - \ln 5 \) ⇒ \( x(\ln 2 - \ln 5) = -\ln 5 - \ln 2 \) ⇒ \( x = \dfrac{\ln 5 + \ln 2}{\ln 5 - \ln 2} = \dfrac{\ln 10}{\ln(5/2)} \).
Solve \( \log_2 (x-1) + \log_2 (x+3) = 3 \).
\( \log_2 ((x-1)(x+3)) = 3 \) ⇒ \( (x-1)(x+3)=8 \) ⇒ \( x^2+2x-3=8 \) ⇒ \( x^2+2x-11=0 \). \( x=-1\pm\sqrt{12}=-1\pm 2\sqrt{3} \). Only \( x=-1+2\sqrt{3} \) (domain \( x>1 \)).