MHB Logarithm Identity: Prove Loga(1/x)=log1/x(a)

[Wild ownz al]

If a>1, a cannot = 1, x>0, show that Log_a(1/x) = log_1/x(a). (COULD NOT SOLVE)

 

[topsquark]

If a>1, a cannot = 1, x>0, show that Log_a(1/x) = log_1/x(a). (COULD NOT SOLVE)

It's no wonder that you can't solve it. It isn't true!

The change of base formula says
[math]log_a(b) = \dfrac{log_c(b)}{log_c(a)}[/math]

Let us change the base of your expressions to, say, base e. Then
[math]log_a \left (\dfrac{1}{x} \right ) = log_{1/x}(a)[/math]

becomes
[math]\dfrac{ln \left ( \dfrac{1}{x} \right ) }{ln(a)} = \dfrac{ln(a)}{ln \left ( \dfrac{1}{x} \right )}[/math]

or
[math]- \dfrac{ln(x)}{ln(a)} = - \dfrac{ln(a)}{ln(x)}[/math]

[math]( ln(a) )^2 = (ln (x) )^2[/math]

So [math]ln(a) = \pm ln(x)[/math]

Clearly this statement isn't true for all x, a.

If the derivation is a bit much, consider the case a = 4, x = 2. Is [math]log_{1/2}(4) = log_4 \left ( \dfrac{1}{2} \right )[/math]?

-Dan