Question

log [1 – {1 – (1 – x2

)–1}–1]–1/2 can be written as

Answer 1

This should be the answer.

Answer 2

Answer:

$\log[1-\{1-(1-x^2)^{-1}\}^{-1}]^{\frac{-1}{2}}]=\log x$

Step-by-step explanation:

Given : Expression $\log[1-\{1-(1-x^2)^{-1}\}^{-1}]^{\frac{-1}{2}}$

To find : Simplify the expression ?

Solution :

Expression $\log[1-\{1-(1-x^2)^{-1}\}^{-1}]^{\frac{-1}{2}}$

$=\log[1-\{1-\frac{1}{1-x^2}\}^{-1}]^{\frac{-1}{2}}$

$=\log[1-\{\frac{1-1-x^2}{1-x^2}\}^{-1}]^{\frac{-1}{2}}$

$=\log[1-\{\frac{-x^2}{1-x^2}\}^{-1}]^{\frac{-1}{2}}$

$=\log[1-\frac{1-x^2}{-x^2}]^{\frac{-1}{2}}$

$=\log[\frac{-x^2-1+x^2}{-x^2}]^{\frac{-1}{2}}$

$=\log[\frac{-1}{-x^2}]^{\frac{-1}{2}}$

$=\log[\frac{1}{x^2}]^{\frac{-1}{2}}$

$=\log[\frac{x^{2\times \frac{1}{2}}}{1}]$

$=\log[x]$

Therefore, $\log[1-\{1-(1-x^2)^{-1}\}^{-1}]^{\frac{-1}{2}}]=\log x$