What is the inverse of g(x)=-4log_{10}(-1/3 (x+2))+1 ?

The inverse of g(x)=-4log_{10}(-1/3 (x+2))+1 is -3*10^{-(x-1)/4}-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of g(x)=-4log_{10}(-1/3 (x+2))+1

inverse

g (x) = - 4 lo g_{10} (- \frac{1}{3} (x + 2)) + 1

- 3 \cdot 1 0^{- \frac{x - 1}{4}} - 2

Solution steps

y = - 4 lo g_{10} (- \frac{1}{3} (x + 2)) + 1

Replace

x

with

y

x = - 4 lo g_{10} (- \frac{1}{3} (y + 2)) + 1

Solve for

y, x = - 4 lo g_{10} (- \frac{1}{3} (y + 2)) + 1

- 3 \cdot 1 0^{- \frac{x - 1}{4}} - 2

What is the inverse of g(x)=-4log_{10}(-1/3 (x+2))+1 ?
The inverse of g(x)=-4log_{10}(-1/3 (x+2))+1 is -3*10^{-(x-1)/4}-2