How do you solve log_12 (p^2-5p)=log_12 (8+2p)?

Hi there! To solve this, you must recognize that if you have logs of the same base, you can drop them, leaving you with the functions inside.

Explanation:

When you have:

${\log}_{b} \left(f \left(x\right)\right) = {\log}_{b} \left(g \left(x\right)\right)$ this is equivalent to:

$f \left(x\right) = g \left(x\right)$

Since both logs in your question have a base of 12, you can drop them, leaving you with:

${p}^{2} - 5 p = 8 + 2 p$

Rearranging we get:

${p}^{2} - 5 p - 2 p - 8 = 0$

Collecting like terms:

${p}^{2} - 7 p - 8 = 0$

Factoring this simple trinomial (finding numbers that multiply to -8 and add to -7):

$\left(p - 8\right) \left(p + 1\right) = 0$

Solving each piece we get:

$p = 8 , - 1$

And that's it, those are the p values that would make those expressions equal. Hopefully everything was clear and helpful! If you have any questions, please feel free to ask! :)

Apr 16, 2016

$p = - 1 \mathmr{and} p = 8$.

Explanation:

Both the logarithms have the same base 12.
So, ${p}^{2} - 5 p = 8 + 2 p$.
${p}^{2} - 7 p - 8 = 0$
The roots of this quadratic equation are$- 1 \mathmr{and} 8.$.
Both the roots are admissible for both the logarithms..