Answer:
[tex]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)[/tex]
Step-by-step explanation:
Given the expression
[tex]3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)[/tex]
solving to write into a single logarithm
[tex]3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)[/tex]
- [tex]\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)[/tex]
[tex]3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)[/tex]
so
[tex]=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)[/tex]
- [tex]\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)[/tex]
[tex]2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)[/tex]
so
[tex]=\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)[/tex]
- [tex]\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)[/tex]
[tex]\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)[/tex]
so
[tex]=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)[/tex]
- [tex]\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)[/tex]
[tex]\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)[/tex]
[tex]=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)[/tex]
Thus,
[tex]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)[/tex]
