dripcima24

2022-10-09

Logarithmic equation help
${\mathrm{log}}_{5}\left(x+3\right)+{\mathrm{log}}_{5}\left(3x-5\right)={\mathrm{log}}_{25}\left(9{x}^{2}\right)$
I have the answer: $\left\{\frac{\sqrt{181}-1}{6}\right\}$ (only answer that falls in the domain)
i understand how to work with the left side but can't figure out how to get the RH side to base 5.

Carson Mueller

Use the fact that $2{\mathrm{log}}_{25}\left(9{x}^{2}\right)={\mathrm{log}}_{5}\left(9{x}^{2}\right)$, which follows from
${5}^{{\mathrm{log}}_{5}\left(9{x}^{2}\right)}=9{x}^{2}={25}^{{\mathrm{log}}_{25}\left(9{x}^{2}\right)}={5}^{2{\mathrm{log}}_{25}\left(9{x}^{2}\right)}.$
Then the equation becomes $2\cdot {\mathrm{log}}_{5}\left(\left(x+3\right)\left(3x-5\right)\right)={\mathrm{log}}_{5}\left(9{x}^{2}\right)$ or ${\mathrm{log}}_{5}\left(\left(\left(x+3\right)\left(3x-5\right){\right)}^{2}\right)={\mathrm{log}}_{5}\left(9{x}^{2}\right)$. This means that $\left(\left(x+3\right)\left(3x-5\right){\right)}^{2}=\left(3x{\right)}^{2}$, hence $\left(x+3\right)\left(3x-5\right)=±3x$. I'm sure you can finish it from here.

charlygyloavao9

We have ${\mathrm{log}}_{25}b=\frac{1}{2}{\mathrm{log}}_{5}b$. For the power to which we must raise $25$ to get $b$ is half the power to which we must raise $5$ to get $b$