Logarithmic equation help

Question

$\log _5\left(x+3\right)+\log _5\left(3x-5\right)=\log _{25}\left(9x^2\right)$

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.

Answer 1

Use the fact that $2\log_{25}(9x^2) = \log_5(9x^2)$, which follows from $$5^{\log_5(9x^2)} = 9x^2 = 25^{\log_{25}(9x^2)} = 5^{2 \log_{25}(9x^2)}.$$

Then the equation becomes $2 \cdot \log_5((x+3)(3x-5)) = \log_5(9x^2)$ or $\log_5(((x+3)(3x-5))^2) = \log_5(9x^2)$. 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.

Answer 2

We have $\log_{25} b=\frac{1}{2}\log_5 b$. For the power to which we must raise $25$ to get $b$ is half the power to which we must raise $5$ to get $b$.

Answer 3

Note that $\log_{25}(9x^2) = 1/2\log_{5}(9x^2) = \log_{5}(3|x|)$. So $(x+3)(3x-5) = 3x^2 + 4x - 15 = 3|x|$ implies $3x^2 + x - 15 = 0$ and $x>0$ or $3x^2 + 7x - 15 = 0$ and $x<0$. However, $3x - 5 > 0$ implies $x > 5/3 > 0$. So we only want the first case. Then you get the desired result.