# How do you solve log_7x-log_7(x-1)=1?

Jul 18, 2016

$x = \frac{7}{6}$

#### Explanation:

When working with logs, the terms must all be in either log form or as numbers, but not a combination of both.

note: ${\log}_{10} 10 = 1 \mathmr{and} {\log}_{2} 2 = 1 \mathmr{and} {\log}_{7} 7 = 1$ etc.#

In order to have all the terms written as logs with the same base, the given equation can be changed to

${\log}_{7} x - {\log}_{7} \left(x - 1\right) = {\log}_{7} 7$

If log terms are added, they can be written as the log of one number:

${\log}_{7} \left(\frac{x}{x - 1}\right) = {\log}_{7} 7$

But if log A = log B, then A = B

So we now have: $\frac{x}{x - 1} = 7$ and can treat it as a normal linear equation:

$x = 7 \left(x - 1\right)$
$x = 7 x - 7$
$7 = 6 x$
$x = \frac{7}{6}$