Since log(ab) = log(a) + log(b) and log(a/b) = log(a) - log(b) for any logarithm base, we have
[tex]\dfrac{\log_x(64) + \log_x(4) - \log_x(8)}{\log_x(1024)} = \dfrac{\log_x\left(\dfrac{64\times4}8\right)}{\log_x(1024)} = \dfrac{\log_x(32)}{\log_x(1024)}[/tex]
Then using the change-of-base identity, we have
[tex]\dfrac{\log_x(32)}{\log_x(1024)} = \log_{1024}(32)[/tex]
But we also know that 2¹⁰ = 1024 and 2⁵ = 32, so we can back up slightly and instead write
[tex]\dfrac{\log_x(32)}{\log_x(1024)} = \dfrac{\log_x(2^5)}{\log_x(2^{10})} = \dfrac{5\log_x(2)}{10\log_x(2)}[/tex]
Then the logarithms cancel and you're left with 5/10 = 1/2.
