# Logarithmic equations with different bases

I had problems understanding how to solve $$6^{-\log_{6}^2}$$

Any help would be much appreciated.

Thanks!

• can you tell me if the expression that you wrote is $\displaystyle 6^{-\log_6 2}$? – DiegoMath Jul 15 '14 at 22:29
• I am guessing that it is. – Juanito Jul 15 '14 at 22:41

Let, $$x=6^{-\log_6 2}$$ Thus, $$\log_6 x =-\log_6 2 = \log_6 2^{-1} = \log_6 \frac{1}{2} \quad \Rightarrow \quad x=\frac{1}{2}$$
$$e^x = y \rightarrow ln(y) = x \\ \rightarrow e^{ln(y)}= e^x=y \\ thus: \\ 6^{-log_6(2)}=({6^{log_6{2}}})^{-1}=(2)^{-1}=1/2$$
\begin{align} 6^{-\log_6(2)} &= \frac{1}{6^{\log_6(2)}} \quad \text{by definition of negative power}\\ &= \frac{1}{2} \quad \text{since 6^x and \log_6(x) are inverses} \end{align}