Welcome to our community

Be a part of something great, join today!

Logarithm Identity

Wild ownz al

Member
Nov 11, 2018
30
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
Aug 30, 2012
1,121
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