What is the inverse of y=sqrt(1-log_{2)(x-1)} ?

The inverse of y=sqrt(1-log_{2)(x-1)} is 2^{1-x^2}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of y=sqrt(1-log_{2)(x-1)}

inverse

y = 1 - lo g_{2} (x - 1)

2^{1 - x^{2}} + 1

Solution steps

y = 1 - lo g_{2} (x - 1)

Replace

x

with

y

x = 1 - lo g_{2} (y - 1)

Solve for

y, x = 1 - lo g_{2} (y - 1)

2^{1 - x^{2}} + 1

f (7) = - 5 x + 2

- \frac{7}{1 + x ^{2}}

f (x) = 11 x^{2} - 4

f (x) = 11 x^{2} - 1

f (x) = 11 x^{2} - 7

What is the inverse of y=sqrt(1-log_{2)(x-1)} ?
The inverse of y=sqrt(1-log_{2)(x-1)} is 2^{1-x^2}+1