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