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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 $$2^x=120$$.
$$\newcommand{\eqncomment}{\small{\text{ #2}} } \newcommand{\ceqns}{\begin{array}{rcll}} \newcommand{\ceqne}{\end{array}}$$
$\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 $$t$$.

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