# Logarithmic and Exponential Equations

## Solving Logarithmic Equations

To solve logarithmic equations we take advantage of the properties of logarithmic functions and the fact that

${\displaystyle y=log_{a}(x){\text{ is equivalent to }}x=a^{y}~a>0,~a\neq 1}$

We also use the additional fact that if ${\displaystyle log_{a}(M)=log_{a}(N)}$ then M = N for M, a, N positive numbers and ${\displaystyle a\neq 1}$

## Example

Solve: ${\displaystyle log_{5}(x+6)+log_{5}(x+2)=1}$

${\displaystyle {\begin{array}{rcl}log_{5}(x+6)+log_{5}(x+2)&=&1\\log_{5}((x+6)(x+2))&=&1\\(x+6)(x+2)&=&5\\x^{2}+8x+12&=&5\\x^{2}+8x+7&=&0\\(x+1)(x+7)&=&0\end{array}}}$

Now we just need to make sure our answers make sense. When x = -7, we have ${\displaystyle log_{5}(-1)+log_{5}(-5)}$ which cannot occur since the domain of the logarithm function is ${\displaystyle (0,\infty )}$

## Solving Exponential Equations

In a similar fashion to solving logarithmic equations, we can solve exponential equations by using their properties and the fact that if ${\displaystyle a^{u}=a^{v}~{\text{ then }}u=v~a>0,~a\neq 1}$

Example:

Solve: ${\displaystyle 8\cdot 3^{x}=5}$

We start by dividing both sides by 8 to get ${\displaystyle 3^{x}={\frac {5}{8}}}$. Taking the log base 3 of both sides we find that ${\displaystyle log_{3}(3^{x})=log_{3}({\frac {5}{8}})}$. Finally by our properties of logarithms ${\displaystyle x=log_{3}({\frac {5}{8}})}$.

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