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

The inverse of f(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

Popular >

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

inverse

f (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

f (x) = 3 lo g_{6} (x)

e^{x^{2} - 2 x}

f (x) = 5 cos (x)

f (x) = \frac{7 x - 2}{4 x + 5}

f (x) = - \frac{3}{8} x^{7}

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