Logarithmic and Exponential Equations

Solving Logarithmic Equations

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

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

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

Example

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

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

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

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 $a^{u}=a^{v}~{\text{ then }}u=v~a>0,~a\neq 1$

Example:

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

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

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