Rewrite both sides with base 2: $8 = 2^3$ and $1/4 = 2^{-2}$ 

So: $(2^3)^{x-1} = (2^{-2})^x$ $\Rightarrow 2^{3(x-1)} = 2^{-2x}$ 

Equate powers: $3(x - 1) = -2x$ 

$3x - 3 = -2x$ 

$5x = 3$ 

$\Rightarrow x = \frac{3}{5}$ 

 We need the value of: 

$\displaystyle \frac{1}{\log_{x+1} 4 - \log_{x+1} 5} + \frac{1}{\log_{1-x} 4 - \log_{1-x} 5}$ 

Use property: $\log_a m - \log_a n = \log_a \left(\frac{m}{n}\right)$ 

So the expression becomes: $\displaystyle \frac{1}{\log_{x+1} \left(\frac{4}{5}\right)} + \frac{1}{\log_{1-x} \left(\frac{4}{5}\right)}$ 

Now use: $\displaystyle \frac{1}{\log_a b} = \log_b a$ 

So expression becomes: $\log_{4/5}(x+1) + \log_{4/5}(1-x)$ 

Use product property: $\log_{4/5}[(x+1)(1-x)]$ 

Compute: $(x+1)(1-x) = 1 - x^2$ 

Substitute $x = \frac{3}{5}$: 

$1 - x^2 = 1 - \frac{9}{25} = \frac{16}{25}$ 

Thus value = $\log_{4/5}\left(\frac{16}{25}\right)$ 

Rewrite: $\frac{16}{25} = \left(\frac{4}{5}\right)^2$ 

Therefore: $\log_{4/5}\left( (4/5)^2 \right) = 2$