# Logarithm Identity

#### Wild ownz al

##### Member
If a>1, a cannot = 1, x>0, show that Loga(1/x) = log1/x(a). (COULD NOT SOLVE)

#### topsquark

##### Well-known member
MHB Math Helper
If a>1, a cannot = 1, x>0, show that Loga(1/x) = log1/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
$$\displaystyle log_a(b) = \dfrac{log_c(b)}{log_c(a)}$$

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