### 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.

\(x={}\) \(16\)

The given equation is: \[\log_{4}(x)+\log_{16}(x)=3\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(4\). So: \[\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(16)}=3\] The left-hand side can be simplified: \[\begin{aligned}\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(4^{2})}&=\log_{4}(x)+\frac{\log_{4}(x)}{2}\\[0.25cm] &=\log_{4}(x)\bigg(1+\frac{1}{2}\bigg)=\frac{3}{2}\log_{4}(x)\\[0.25cm]\end{aligned}\] So \[\frac{3}{2}\log_{4}(x)=3\] Thus: \[\log_{4}(x)=2\] On the basis of the definition of the logarithm with base 4 we get: \[\begin{aligned}x&=4^{2}\\[0.25cm] &=16\end{aligned}\]

The given equation is: \[\log_{4}(x)+\log_{16}(x)=3\] The first step is to equalize the bases. It doesn't matter what base you take. Here we set the bases equal to \(4\). So: \[\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(16)}=3\] The left-hand side can be simplified: \[\begin{aligned}\log_{4}(x)+\frac{\log_{4}(x)}{\log_{4}(4^{2})}&=\log_{4}(x)+\frac{\log_{4}(x)}{2}\\[0.25cm] &=\log_{4}(x)\bigg(1+\frac{1}{2}\bigg)=\frac{3}{2}\log_{4}(x)\\[0.25cm]\end{aligned}\] So \[\frac{3}{2}\log_{4}(x)=3\] Thus: \[\log_{4}(x)=2\] On the basis of the definition of the logarithm with base 4 we get: \[\begin{aligned}x&=4^{2}\\[0.25cm] &=16\end{aligned}\]

Unlock full access