Answer:
x = 3
Step-by-step explanation:
Given:[tex]\log _{x-1}\left(16\right)=4[/tex]
We have to solve for x.
Consider the given expression [tex]\log _{x-1}\left(16\right)=4[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _a\left(b\right)=\frac{\ln \left(b\right)}{\ln \left(a\right)}[/tex]
[tex]\log _{x-1}\left(16\right)=\frac{\ln \left(16\right)}{\ln \left(x-1\right)}[/tex]
Thus, the expression becomes,
[tex]\frac{\ln \left(16\right)}{\ln \left(x-1\right)}=4[/tex]
Multiply both side by [tex]\ln \left(x-1\right)[/tex]
[tex]\frac{\ln \left(16\right)}{\ln \left(x-1\right)}\ln \left(x-1\right)=4\ln \left(x-1\right)[/tex]
Simplify, we get,
[tex]\ln \left(16\right)=4\ln \left(x-1\right)[/tex]
Divide both side by 4, we have,
[tex]\frac{4\ln \left(x-1\right)}{4}=\frac{\ln \left(16\right)}{4}[/tex]
[tex]\frac{\ln \left(16\right)}{4}:\quad \ln \left(2\right)[/tex]
Thus, [tex]\ln \left(x-1\right)=\ln(2)[/tex]
[tex]\mathrm{When\:the\:logs\:have\:the\:same\:base:\:\:}\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)[/tex]
thus, x - 1 = 2
simplify for x, we have,
x = 3
