Bradyn Mclean

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

Beginner2023-01-01Added 7 answers

You must employ the following logarithm properties in order to make this expression simpler:

$\mathrm{log}(a\cdot b)=\mathrm{log}\left(a\right)+\mathrm{log}\left(b\right)$ (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:

${\mathrm{log}}_{a}\left(\frac{{x}^{\frac{1}{2}}{y}^{4}}{{x}^{3}}\right)={\mathrm{log}}_{a}\left({x}^{-\frac{5}{2}}{y}^{4}\right)$

Using the property (3), you have:

Then, using the properties (1) and (2), you have:

Then, you only need to put all the powers of

together:

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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