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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 Cedie Camomot
1 year, 8 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:- \[\log_b a^x=\log_b n\] \[\implies x \log_b a=\log_b n \] \[\implies x=\dfrac{\log_b n}{\log_b a}\]

Rishabh Cool - 1 year, 8 months ago

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\[\begin{equation} \begin{split} x & =\log_{a}n = \frac{\log n}{\log a} \\ & = \frac{ \frac{\log n}{\log b}}{\frac{\log a}{\log b}} = \frac{\log_{b} n}{\log_{b}a}\end{split} \end{equation}\]

Akshat Sharda - 1 year, 8 months ago

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

Vishnu Bhagyanath - 1 year, 8 months ago

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