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#### Wild ownz al

##### Member

- Nov 11, 2018

- 30

If a>1, a cannot = 1, x>0, show that Log

_{a}(1/x) = log_{1/x}(a). (COULD NOT SOLVE)- Thread starter Wild ownz al
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- Thread starter
- #1

- Nov 11, 2018

- 30

If a>1, a cannot = 1, x>0, show that Log_{a}(1/x) = log_{1/x}(a). (COULD NOT SOLVE)

- Aug 30, 2012

- 1,120

It's no wonder that you can't solve it. It isn't true!If a>1, a cannot = 1, x>0, show that Log_{a}(1/x) = log_{1/x}(a). (COULD NOT SOLVE)

The change of base formula says

\(\displaystyle log_a(b) = \dfrac{log_c(b)}{log_c(a)}\)

Let us change the base of your expressions to, say, base e. Then

\(\displaystyle log_a \left (\dfrac{1}{x} \right ) = log_{1/x}(a)\)

becomes

\(\displaystyle \dfrac{ln \left ( \dfrac{1}{x} \right ) }{ln(a)} = \dfrac{ln(a)}{ln \left ( \dfrac{1}{x} \right )}\)

or

\(\displaystyle - \dfrac{ln(x)}{ln(a)} = - \dfrac{ln(a)}{ln(x)}\)

\(\displaystyle ( ln(a) )^2 = (ln (x) )^2\)

So \(\displaystyle ln(a) = \pm ln(x)\)

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 \(\displaystyle log_{1/2}(4) = log_4 \left ( \dfrac{1}{2} \right )\)?

-Dan