Exponential functions and logarithms: Logarithms
Equations and inequalities containing logarithms with different bases
When logarithms with different bases are present in an equation or inequality, the calculation rule \[\log_g(x) =\frac{\log_h(x)}{\log_h(g)}\] helps to convert all logarithms into logarithms with the same base.
The examples below show how it works.
\(x={}\) \(27\)
The given equation is: \[\log_{3}(x)+\log_{27}(x)=4\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(3\). So: \[\log_{3}(x)+\frac{\log_{3}(x)}{\log_{3}(27)}=4\] The left-hand side can be simplified: \[\begin{aligned}\log_{3}(x)+\frac{\log_{3}(x)}{\log_{3}(3^{3})}&=\log_{3}(x)+\frac{\log_{3}(x)}{3}\\[0.25cm] &=\log_{3}(x)\bigg(1+\frac{1}{3}\bigg)=\frac{4}{3}\log_{3}(x)\\[0.25cm]\end{aligned}\] So \[\frac{4}{3}\log_{3}(x)=4\] Thus: \[\log_{3}(x)=3\] On the basis of the definition of the logarithm with base 3 we get: \[\begin{aligned}x&=3^{3}\\[0.25cm] &=27\end{aligned}\]
The given equation is: \[\log_{3}(x)+\log_{27}(x)=4\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(3\). So: \[\log_{3}(x)+\frac{\log_{3}(x)}{\log_{3}(27)}=4\] The left-hand side can be simplified: \[\begin{aligned}\log_{3}(x)+\frac{\log_{3}(x)}{\log_{3}(3^{3})}&=\log_{3}(x)+\frac{\log_{3}(x)}{3}\\[0.25cm] &=\log_{3}(x)\bigg(1+\frac{1}{3}\bigg)=\frac{4}{3}\log_{3}(x)\\[0.25cm]\end{aligned}\] So \[\frac{4}{3}\log_{3}(x)=4\] Thus: \[\log_{3}(x)=3\] On the basis of the definition of the logarithm with base 3 we get: \[\begin{aligned}x&=3^{3}\\[0.25cm] &=27\end{aligned}\]
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