Introduction to Logarithms
Using logarithms to solve equations
- In any equation, if the unknown is in the power (exponent) then logarithms can be used to solve the equation.
- Remember to follow the rules and principles of algebra.
For example: Solve \(2^x=120\).
\(
\newcommand{\eqncomment}[2]{\small{\text{ #2}} }
\newcommand{\ceqns}{\begin{array}{rcll}}
\newcommand{\ceqne}{\end{array}}
\)
\[
\ceqns
2^x &=& 120 \\
\log 2^x &=& \log 120 & \eqncomment{0.5}{take the logarithm of both sides} \\
x \log 2 &=& \log 120 & \eqncomment {0.5}{using the logarithm of a power rule}\\
\frac{x\log 2}{\log 2} &=& \frac{\log 120}{\log 2} & \eqncomment{0.5}{dividing both sides by \(\log 2\)}\\
x &=& \frac{\log 120}{\log 2} &\eqncomment{0.5}{evaluating on the calculator}\\
&\approx& 6.907
\ceqne
\]
Another example would be to solve \(100 = 50(10)^{3t}\) for \(t\).
\[
\ceqns
100 &=& 50(10)^{3t} \\
\frac{100}{50} &=& 10^{3t} & \eqncomment{0.5}{dividing both sides by 50}\\
2 &=& 10^{3t} \\
\log 2 &=& \log 10^{3t} & \eqncomment{0.5}{take the logarithm of both sides} \\
\log 2 &=& 3t \log10 & \eqncomment {0.5}{using the logarithm of a power rule}\\
\log 2 &=& 3t \\
\frac{\log 2}{3} &=& t \\
t &\approx& 0.1003
\ceqne
\]
To do
- Solve exponential equations using logarithms: base-10 and base-\(e\) activity (Khan Academy)
- Solve exponential equations using logarithms: base-2 and other bases activity (Khan Academy)
More info
- Introduction to logarithms video(Study Support, USQ Library)
- Solving logarithmic equations video (Study Support, USQ Library)
- Solving equations using logs Quick reference guide (mathcentre)
- Logarithmic equations: variable in the base video (Khan Academy)
- Solving logarithmic equations video (Khan Academy)