Answer:
Assuming log is the natural logarithm:
[tex]f^{-1}(x)=\frac{e^{x/2}}{3}[/tex]
Assuming log is the base 10 logarithm:
[tex]f^{-1}(x)=\frac{10^{x/2}}{3}[/tex]
Step-by-step explanation:
Assuming log is the natural logarithm:
First step: Replace f(x) with y that is [tex]f(x)=y[/tex]
[tex]y=2log(3x)[/tex]
Second step: Solve for x:
[tex]y=2log(3x)\\\\\frac{y}{2} =log(3x)[/tex]
[tex]e^{\frac{y}{2} }=e^{log(3x)}\\\\e^{\frac{y}{2} }=3x\\\\\frac{e^{\frac{y}{2} }}{3} =x[/tex]
Third step: Replace every x with a y and replace every y with an x:
[tex]\frac{e^{\frac{x}{2} }}{3} =y[/tex]
Final step: Replace y with [tex]f^{-1}(x)[/tex]:
[tex]f^{-1}(x)=\frac{e^{x/2}}{3}[/tex]
Assuming log is the base 10 logarithm:
It is the same procedure as before, the only thing that change, is when we are solving for x:
First step: Replace f(x) with y that is [tex]f(x)=y[/tex]
[tex]y=2log(3x)[/tex]
Second step: Solve for x:
[tex]y=2log(3x)\\\\\frac{y}{2} =log(3x)[/tex]
[tex]10^{\frac{y}{2} }=10^{log(3x)}\\\\10^{\frac{y}{2} }=3x\\\\\frac{10^{\frac{y}{2} }}{3} =x[/tex]
Third step: Replace every x with a y and replace every y with an x:
[tex]\frac{10^{\frac{x}{2} }}{3} =y[/tex]
Final step: Replace y with [tex]f^{-1}(x)[/tex] :
[tex]f^{-1}(x)=\frac{10^{x/2}}{3}[/tex]
