Answer:
Look at the picture
Step-by-step explanation:
[tex]\log_ab=c\iff a^c=b\\\\\text{Therefore}\\\\f(x)=b^x\to y=b^x\\\\\text{find}\ f^{-1}(x).\ \text{Exchange}\ x\ \text{and}\ y\ \text{and solve for}\ y:\\\\b^y=x\qquad\text{logarithm iwth base b of both sides}\\\\\log_bb^y=\log_b(x)\qquad\text{use}\ \log_ab^m=m\log_ab\\\\y\log_bb=\log_b(x)\qquad\text{use}\ \log_aa=1\\\\y=\log_b(x)\\\\f^{-1}(x)=\log_b(x)\to g(x)=f^{-1}(x)\\\\\text{Then:}\ f(a)=b\to g(b)=a[/tex]
