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Prove this, where a > 0, a cannot equal 1, x > 0
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\[\log_{a}(1/x) = \log_{1/a}(x) \]
didn't like the last one? or was it confusing?
we can prove it a different way if you like
put \[\log_a(\frac{1}{x})=y\] and \[\log_{\frac{1}{a}}(x)=z\] and we want to show that \(y=z\)
writing in equivalent exponential form give \[\log_a(\frac{1}{x})=y\iff a^y=\frac{1}{x}\] and \[\log_{\frac{1}{a}}(x)=z\iff (\frac{1}{a})^z=x\] therefore \[a^{-y}=x\] and also \[a^{-z}=x\] and since exponentials are one to one functions this implies that \[-y=-z\] and so \[y=z\] it is really the same as the last one, just looks different
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No i got the last one, wanted a fast bump, but you answered the previous one.
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