Respuesta :

[tex]\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}[/tex]

[tex]\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[/tex]

Answer:

[tex]x=\frac{log(15)}{log(3)}-1[/tex]

Step-by-step explanation:

[tex]3^{x+1} = 15[/tex]

LEts convert exponential form to log form

[tex]b^x=a[/tex] can be written as [tex]log_b(a)=x[/tex]

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

[tex]3^{x+1} = 15[/tex]

[tex]log_3{15} = x+1[/tex]

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

[tex]log_b(a)=\frac{log(a)}{log(b)}[/tex]

[tex]log_3{15} = x+1[/tex]

[tex]\frac{log(15)}{log(3)} = x+1[/tex]

Now subtract 1 from both sides

[tex]x=\frac{log(15)}{log(3)}-1[/tex]