Aspire Faculty ID #11946 · Topic: NIMCET 2025 · Just now
NIMCET 2025

If $8^{x-1}=(1/4)^{x}$, then the value of $\frac{1}{\log_{x+1}4-\log_{x+1}5}+\frac{1}{\log_{1-x}4-\log_{1-x}5}$ is

Solution

Given: $8^{x-1} = (1/4)^{x}$ 
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: $\log_{4/5}\left( (4/5)^2 \right) = 2$
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