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.
\(x={}\) \(64\)
The given equation is: \[\log_{4}(x)+\log_{64}(x)=4\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(4\). So: \[\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(64)}=4\] The left-hand side can be simplified: \[\begin{aligned}\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(4^{3})}&=\log_{4}(x)+\frac{\log_{4}(x)}{3}\\[0.25cm] &=\log_{4}(x)\bigg(1+\frac{1}{3}\bigg)=\frac{4}{3}\log_{4}(x)\\[0.25cm]\end{aligned}\] So \[\frac{4}{3}\log_{4}(x)=4\] Thus: \[\log_{4}(x)=3\] On the basis of the definition of the logarithm with base 4 we get: \[\begin{aligned}x&=4^{3}\\[0.25cm] &=64\end{aligned}\]
The given equation is: \[\log_{4}(x)+\log_{64}(x)=4\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(4\). So: \[\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(64)}=4\] The left-hand side can be simplified: \[\begin{aligned}\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(4^{3})}&=\log_{4}(x)+\frac{\log_{4}(x)}{3}\\[0.25cm] &=\log_{4}(x)\bigg(1+\frac{1}{3}\bigg)=\frac{4}{3}\log_{4}(x)\\[0.25cm]\end{aligned}\] So \[\frac{4}{3}\log_{4}(x)=4\] Thus: \[\log_{4}(x)=3\] On the basis of the definition of the logarithm with base 4 we get: \[\begin{aligned}x&=4^{3}\\[0.25cm] &=64\end{aligned}\]
Unlock full access
