$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

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3 years ago