If \(8^{x-1} = (\frac{1}{4})^x\), 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
Step-by-step Solution:
Correct Answer: A (2)
Note on the problem statement: The question appears to contain two typos.
We start with the corrected equation and express both sides with a common base of 2.
$$ 8^{z-1} = \left(\frac{1}{4}\right)^z $$
$$ (2^3)^{z-1} = (2^{-2})^z $$
$$ 2^{3(z-1)} = 2^{-2z} $$
Since the bases are equal, we can equate the exponents:
$3(z-1) = -2z$
$3z - 3 = -2z$
$5z = 3$
$z = \frac{3}{5}$
The original expression has an undefined term $\log_{z-1}$ because $z-1 = 3/5 - 1 = -2/5$. We assume the expression was meant to be:
$$ \frac{1}{\log_{z+1} 4 - \log_{z+1} 5} + \frac{1}{\log_{1-z} 4 - \log_{1-z} 5} $$
First, we simplify the denominators using the logarithm property $\log_b M - \log_b N = \log_b(\frac{M}{N})$:
$$ \frac{1}{\log_{z+1}(4/5)} + \frac{1}{\log_{1-z}(4/5)} $$
Next, we use the change of base property $\frac{1}{\log_b a} = \log_a b$:
$$ \log_{4/5}(z+1) + \log_{4/5}(1-z) $$
Finally, we combine the terms using the property $\log_b M + \log_b N = \log_b(MN)$:
$$ \log_{4/5}((z+1)(1-z)) = \log_{4/5}(1-z^2) $$
Now we substitute our calculated value $z = \frac{3}{5}$ into the simplified expression:
$$ \log_{4/5}\left(1 - \left(\frac{3}{5}\right)^2\right) $$
$$ = \log_{4/5}\left(1 - \frac{9}{25}\right) $$
$$ = \log_{4/5}\left(\frac{16}{25}\right) $$
We can rewrite the argument as a power of the base:
$$ = \log_{4/5}\left(\left(\frac{4}{5}\right)^2\right) $$
Using the property $\log_b(b^p) = p$, we get the final answer:
$$ = 2 $$