# What is the inverse of y= -log (1.05x+10^-2) ?

Jun 25, 2017

${f}^{-} 1 \left(x\right) = \frac{{10}^{-} x - {10}^{-} 2}{1.05}$

#### Explanation:

Given: $f \left(x\right) = - \log \left(1.05 x + {10}^{-} 2\right)$

Let $x = {f}^{-} 1 \left(x\right)$

$f \left({f}^{-} 1 \left(x\right)\right) = - \log \left(1.05 {f}^{-} 1 \left(x\right) + {10}^{-} 2\right)$

By definition $f \left({f}^{-} 1 \left(x\right)\right) = x$

$x = - \log \left(1.05 {f}^{-} 1 \left(x\right) + {10}^{-} 2\right)$

Multiply both sides by -1:

$- x = \log \left(1.05 {f}^{-} 1 \left(x\right) + {10}^{-} 2\right)$

Make both sides the exponent of 10:

${10}^{-} x = {10}^{\log \left(1.05 {f}^{-} 1 \left(x\right) + {10}^{-} 2\right)}$

Because 10 and log are inverses, the right side reduces to the argument:

${10}^{-} x = 1.05 {f}^{-} 1 \left(x\right) + {10}^{-} 2$

Flip the equation:

$1.05 {f}^{-} 1 \left(x\right) + {10}^{-} 2 = {10}^{-} x$

Subtract 10^-2 from both sides:

$1.05 {f}^{-} 1 \left(x\right) = {10}^{-} x - {10}^{-} 2$

Divide both sides by 1.05:

${f}^{-} 1 \left(x\right) = \frac{{10}^{-} x - {10}^{-} 2}{1.05}$

Check:

$f \left({f}^{-} 1 \left(x\right)\right) = - \log \left(1.05 \left(\frac{{10}^{-} x - {10}^{-} 2}{1.05}\right) + {10}^{-} 2\right)$

$f \left({f}^{-} 1 \left(x\right)\right) = - \log \left({10}^{-} x - {10}^{-} 2 + {10}^{-} 2\right)$

$f \left({f}^{-} 1 \left(x\right)\right) = - \log \left({10}^{-} x\right)$

$f \left({f}^{-} 1 \left(x\right)\right) = - \left(- x\right)$

$f \left({f}^{-} 1 \left(x\right)\right) = x$

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{{10}^{-} \left(- \log \left(1.05 x + {10}^{-} 2\right)\right) - {10}^{-} 2}{1.05}$

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{{10}^{\log \left(1.05 x + {10}^{-} 2\right)} - {10}^{-} 2}{1.05}$

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{1.05 x + {10}^{-} 2 - {10}^{-} 2}{1.05}$

${f}^{-} 1 \left(f \left(x\right)\right) = \frac{1.05 x}{1.05}$

${f}^{-} 1 \left(f \left(x\right)\right) = x$

Both conditions check.