Question

Simplify without using a calculator :

[log_{x}(64) + log_{x}(4) - log_{x}(8)] ÷[log_{x}(1024)]
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Answer 1

Since log(ab) = log(a) + log(b) and log(a/b) = log(a) - log(b) for any logarithm base, we have

`(\log_x(64) + \log_x(4) - \log_x(8))/(\log_x(1024)) = \frac{\log_x\left(\frac{64*4}8\right)}{\log_x(1024)} = (\log_x(32))/(\log_x(1024))`

Then using the change-of-base identity, we have

`(\log_x(32))/(\log_x(1024)) = \log_(1024)(32)`

But we also know that 2¹⁰ = 1024 and 2⁵ = 32, so we can back up slightly and instead write

`(\log_x(32))/(\log_x(1024)) = (\log_x(2^5))/(\log_x(2^(10))) = (5\log_x(2))/(10\log_x(2))`

Then the logarithms cancel and you're left with 5/10 = 1/2.