dripcima24

2022-10-09

Logarithmic equation help

${\mathrm{log}}_{5}(x+3)+{\mathrm{log}}_{5}(3x-5)={\mathrm{log}}_{25}\left(9{x}^{2}\right)$

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.

${\mathrm{log}}_{5}(x+3)+{\mathrm{log}}_{5}(3x-5)={\mathrm{log}}_{25}\left(9{x}^{2}\right)$

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

Beginner2022-10-10Added 7 answers

Use the fact that $2{\mathrm{log}}_{25}(9{x}^{2})={\mathrm{log}}_{5}(9{x}^{2})$, which follows from

${5}^{{\mathrm{log}}_{5}(9{x}^{2})}=9{x}^{2}={25}^{{\mathrm{log}}_{25}(9{x}^{2})}={5}^{2{\mathrm{log}}_{25}(9{x}^{2})}.$

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

${5}^{{\mathrm{log}}_{5}(9{x}^{2})}=9{x}^{2}={25}^{{\mathrm{log}}_{25}(9{x}^{2})}={5}^{2{\mathrm{log}}_{25}(9{x}^{2})}.$

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

charlygyloavao9

Beginner2022-10-11Added 2 answers

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