# Logarithms

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Exponential relations can also be expressed via logarithms. For example, the fact $${2^3} = 8$$ can be expressed as: $${\log _2}8 = 3$$ . This is read as follows: the logarithm of 8 to base 2 is 3. Thus, the logarithm of a number to some base tells us the power to which the base must be raised to obtain that number. Some more examples:

\begin{align}&3^4=81\qquad\quad\Rightarrow\qquad\log_381=4\\&5^{-2}=\frac1{25}\quad\quad\Rightarrow\qquad \log_5\frac1{25}=-2\\&16\frac14=2\quad\quad\;\;\Rightarrow\qquad\log_{16}2=\frac14\\&\left(\frac12\right)^5=\frac1{32}\quad\Rightarrow\qquad\log_{\left(\frac12\right)}\left(\frac1{32}\right)=5\\\end{align}

In general,${b^e=N\;\;\;\;\;\;\;\;\Rightarrow\;\;\;\;\;\;\log_bN=e}$

Or, in words:

\begin{align}&{\rm{Bas}}{{\rm{e}}^{{\rm{Exponent}}}} = {\rm{Number}}\\& \Rightarrow \,\,{\log _{{\rm{Base}}}}{\rm{Number}} = {\rm{Exponent}}\end{align}

Let us calculate $${\log _{2\sqrt 2 }}\left( {32 \times {4^{\frac{1}{5}}}} \right)$$. We have:

\begin{align}&{\log _{2\sqrt 2 }}\left( {32 \times {4^{\frac{1}{5}}}} \right) = {\log _{\left( {{2^{\frac{3}{2}}}} \right)}}\left( {{2^5} \times {2^{\frac{2}{5}}}} \right)\\ &= {\log _{\left( {{2^{\frac{3}{2}}}} \right)}}\left( {{2^{5 + \frac{2}{5}}}} \right) = {\log _{\left( {{2^{\frac{3}{2}}}} \right)}}\left( {{2^{\frac{{27}}{5}}}} \right)\end{align}

This logarithm is a number e such that

\begin{align} & {\left( {{2^{\frac{3}{2}}}} \right)^e} = \left( {{2^{\frac{{27}}{5}}}} \right)\\& \Rightarrow \,\,\,\frac{3}{2} \times e = \frac{{27}}{5}\,\,\, \Rightarrow \,\,\,e = \frac{{18}}{5}\,\,\,{\rm{or}}\,\,\,3.6\end{align}

Thus,

${\log _{2\sqrt 2 }}\left( {32 \times {4^{\frac{1}{5}}}} \right) = \frac{{18}}{5}$