How to solve for log with a number outside?

Question

$$\log_6(4x-10)+1 = \log_6(15x+15)$$

This is a sample problem. I know that when the bases of log are the same, all you have to do is set the parenthesis inside equal to each other.

If the $1$ wasn't there, I would just do $4x -10 = 15x + 15$. However, I don't know what to do with that $1$.

Answer 1

Write $1 = \log_6 6$ then your equation reduces to $\log_6 (4x-10) + \log_6 6 = \log_6 (15x+15)$ upon which, using the logarithm addition law ($\log a + \log b = \log ab$) and injectivity of the logarithm, we have:

$$6(4x-10) = 15x+15 \iff 9x = 75 \iff x = \frac{25}{3}$$

In general, you can always write a real number $a$ as:

$$a = a \log_b b = \log_b b^a$$

Answer 2

You can always replace a real number, say $x$, with a logarithmic number like $\log_b{b^x} $. Here, as Zain did, $1=\log_66$.