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.
\(x={}\) \(25\)
The given equation is: \[\log_{5}(x)+\log_{25}(x)=3\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(5\). So: \[\log_{5}(x)+\frac{\log_{5}(x)}{\log_{5}(25)}=3\] The left-hand side can be simplified: \[\begin{aligned}\log_{5}(x)+\frac{\log_{5}(x)}{\log_{5}(5^{2})}&=\log_{5}(x)+\frac{\log_{5}(x)}{2}\\[0.25cm] &=\log_{5}(x)\bigg(1+\frac{1}{2}\bigg)=\frac{3}{2}\log_{5}(x)\\[0.25cm]\end{aligned}\] So \[\frac{3}{2}\log_{5}(x)=3\] Thus: \[\log_{5}(x)=2\] On the basis of the definition of the logarithm with base 5 we get: \[\begin{aligned}x&=5^{2}\\[0.25cm] &=25\end{aligned}\]
The given equation is: \[\log_{5}(x)+\log_{25}(x)=3\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(5\). So: \[\log_{5}(x)+\frac{\log_{5}(x)}{\log_{5}(25)}=3\] The left-hand side can be simplified: \[\begin{aligned}\log_{5}(x)+\frac{\log_{5}(x)}{\log_{5}(5^{2})}&=\log_{5}(x)+\frac{\log_{5}(x)}{2}\\[0.25cm] &=\log_{5}(x)\bigg(1+\frac{1}{2}\bigg)=\frac{3}{2}\log_{5}(x)\\[0.25cm]\end{aligned}\] So \[\frac{3}{2}\log_{5}(x)=3\] Thus: \[\log_{5}(x)=2\] On the basis of the definition of the logarithm with base 5 we get: \[\begin{aligned}x&=5^{2}\\[0.25cm] &=25\end{aligned}\]
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