Answer:
The inverse of the function [tex]y=3e^{4x+1}[/tex] is [tex]f^{-1}(x) =\frac{\ln \left(\frac{x}{3}\right)-1}{4}[/tex]
Step-by-step explanation:
Given the function [tex]y=3e^{4x+1}[/tex] we want to find the inverse function, [tex]f^{-1}(x)[/tex]
Applying the above process we get:
[tex]\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y\\\\x=3e^{4y+1}\\\\\mathrm{Solve}\:x=3e^{4y+1}\:\mathrm{for}\:y\\\\3e^{4y+1}=x\\\\\frac{3e^{4y+1}}{3}=\frac{x}{3}\\\\e^{4y+1}=\frac{x}{3}[/tex]
[tex]\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)\\\\\ln \left(e^{4y+1}\right)=\ln \left(\frac{x}{3}\right)\\\\\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\\\\\left(4y+1\right)\ln \left(e\right)=\ln \left(\frac{x}{3}\right)\\\\4y+1=\ln \left(\frac{x}{3}\right)\\\\y=\frac{\ln \left(\frac{x}{3}\right)-1}{4}\\\\f^{-1}(x) =\frac{\ln \left(\frac{x}{3}\right)-1}{4}[/tex]
The inverse of the function [tex]y=3e^{4x+1}[/tex] is [tex]f^{-1}(x) =\frac{\ln \left(\frac{x}{3}\right)-1}{4}[/tex]
