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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 \(t\).

\[
\ceqns 
100 &=& 50(10)^{3t} \\  
\frac{100}{50} &=& 10^{3t} & \eqncomment{0.5}{dividing both sides by 50}\\  
2 &=& 10^{3t} \\  
\log 2 &=& \log 10^{3t} & \eqncomment{0.5}{take the logarithm of both sides} \\  
\log 2 &=& 3t \log10 & \eqncomment {0.5}{using the logarithm of a power rule}\\  
\log 2 &=& 3t \\  
\frac{\log 2}{3} &=& t \\  
t &\approx& 0.1003
\ceqne
\]