# Logarithm problem 2

Let $$x$$ be any real number and let $$a, b, n$$ be positive real number with $$a \not = 1$$ and $$b \not = 1$$ . Show that if $$x = \log_{a}n$$ then $$x = \displaystyle{\frac{\log_{b}n}{\log_{b}a}}$$.

Note by A Brilliant Member
2 years, 4 months ago

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By definition of logs:- $x = \log_{a}n\implies a^x=n$ Taking $$log$$ with base $$b$$ on both sides:- $\log_b a^x=\log_b n$ $\implies x \log_b a=\log_b n$ $\implies x=\dfrac{\log_b n}{\log_b a}$

- 2 years, 4 months ago

$$$\begin{split} x & =\log_{a}n = \frac{\log n}{\log a} \\ & = \frac{ \frac{\log n}{\log b}}{\frac{\log a}{\log b}} = \frac{\log_{b} n}{\log_{b}a}\end{split}$$$

- 2 years, 4 months ago

Well that's basically applying what we need to prove with b=e.

- 2 years, 4 months ago