What is the inverse of log_{x}(8)= I/(1*10^{-12)} ?

The inverse of log_{x}(8)= I/(1*10^{-12)} is 2^{3/(x*10^{12)}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of log_{x}(8)= I/(1*10^{-12)}

inverse

lo g_{x} (8) = \frac{I}{1 \cdot 1 0 ^{- 12}}

2^{\frac{3}{x \cdot 1 0 ^{12}}}

Solution steps

I = \frac{3}{1 0 ^{12}} lo g_{x} (2)

Replace

x

with

I

x = \frac{3}{1 0 ^{12}} lo g_{I} (2)

Solve for

I, x = \frac{3}{1 0 ^{12}} lo g_{I} (2)

2^{\frac{3}{x \cdot 1 0 ^{12}}}

f (x) = \frac{( - x - 1 )}{2 x - 1}

17 m^{2} + 50 m + 165

f (x) = \frac{1}{2} (x - 1) (x - 1) - 2

f (x) = - 3 2 x - 1 - 4

What is the inverse of log_{x}(8)= I/(1*10^{-12)} ?
The inverse of log_{x}(8)= I/(1*10^{-12)} is 2^{3/(x*10^{12)}}