 2022-12-31

How to express as a single logarithm & simplify $\left(\frac{1}{2}\right){\mathrm{log}}_{a}\cdot x+4{\mathrm{log}}_{a}\cdot y-3{\mathrm{log}}_{a}\cdot x$? possetzvjm

Expert

You must employ the following logarithm properties in order to make this expression simpler:
$\mathrm{log}\left(a\cdot b\right)=\mathrm{log}\left(a\right)+\mathrm{log}\left(b\right)$ (1)
$\mathrm{log}\left(\frac{a}{b}\right)=\mathrm{log}\left(a\right)-\mathrm{log}\left(b\right)$ (2)
$\mathrm{log}\left({a}^{b}\right)=b\mathrm{log}\left(a\right)$ (3)
Using the property (3), you have:
$\left(\frac{1}{2}\right){\mathrm{log}}_{a}\left(x\right)+4{\mathrm{log}}_{a}\left(y\right)-3{\mathrm{log}}_{a}\left(x\right)={\mathrm{log}}_{a}\left({x}^{\frac{1}{2}}\right)+{\mathrm{log}}_{a}\left({y}^{4}\right)-{\mathrm{log}}_{a}\left({x}^{3}\right)$
Then, using the properties (1) and (2), you have:
${\mathrm{log}}_{a}\left({x}^{\frac{1}{2}}\right)+{\mathrm{log}}_{a}\left({y}^{4}\right)-{\mathrm{log}}_{a}\left({x}^{3}\right)={\mathrm{log}}_{a}\left(\frac{{x}^{\frac{1}{2}}{y}^{4}}{{x}^{3}}\right)$
Then, you only need to put all the powers of $x$
together:
${\mathrm{log}}_{a}\left(\frac{{x}^{\frac{1}{2}}{y}^{4}}{{x}^{3}}\right)={\mathrm{log}}_{a}\left({x}^{-\frac{5}{2}}{y}^{4}\right)$

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