Question

Solve 3^(x+1) = 15 for x using the change of base formula

Answer 1

`\bf \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \\\\\\ \textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\bf 3^(x+1)=15\implies \log_(10)(3^(x+1))=\log_(10)(15)\implies (x+1)\log_(10)(3)=\log_(10)(15) \\\\\\ x+1=\cfrac{\log_(10)(15)}{\log_(10)(3)}\implies \stackrel{\textit{change of base rule}}{x=\cfrac{\log_(e)(15)}{\log_(e)(3)}-1}\implies x\approx 1.47`

Answer 2

`x=(log(15))/(log(3))-1`

Explanation:

`3^(x+1) = 15`

LEts convert exponential form to log form

`b^x=a` can be written as
`log_b(a)=x`

WE apply the same rule to convert the given exponential form to log form

`3^(x+1) = 15`

`log_3{15} = x+1`

HEre the base of log is 3. Lets apply change of base formula

`log_b(a)=(log(a))/(log(b))`

`log_3{15} = x+1`

`(log(15))/(log(3)) = x+1`

Now subtract 1 from both sides

`x=(log(15))/(log(3))-1`