Respuesta :

frika

Answer:

[tex]f^{-1}(x)=\dfrac{e^x}{3},\ x\in(-\infty,\infty),\ y>0[/tex]

Step-by-step explanation:

Consider the function [tex]f(x)=\ln 3x.[/tex] the domain of this function is x>0 and the range of this function is all real numbers. The inverse function has the domain all real numbers and the range y>0.

If

[tex]y=\ln 3x,[/tex]

then

[tex]3x=e^y,\\ \\x=\dfrac{e^{y}}{3}.[/tex]

Change x into y and y into x:

[tex]y=\dfrac{e^x}{3}.[/tex]

Thus, the inverse function is

[tex]f^{-1}(x)=\dfrac{e^x}{3}.[/tex]

kudzordzifrancis

ANSWER

[tex]{f}^{ - 1} (x) = \frac{ {e}^{x} }{3} [/tex]

EXPLANATION

The given logarithmic function is

[tex]f(x) = ln(3x) [/tex]

Let

[tex]y = ln(3x) [/tex]

Interchange x and y.

[tex]x = ln(3y) [/tex]

Solve for y.

[tex] {e}^{x} = {e}^{ ln(3y) } [/tex]

[tex] {e}^{x} = 3y[/tex]

Divide both sides by 3.

[tex] \frac{ {e}^{x} }{3} = y[/tex]

Therefore

[tex] {f}^{ - 1} (x) = \frac{ {e}^{x} }{3} [/tex]

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