# How do you solve Log(2x+1)=6+log(x-1)?

Feb 12, 2016

Since the logs are in a common base we can put all logs to one side an simplify by using the logarithm law ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$

#### Explanation:

$\log \left(2 x + 1\right) - \log \left(x - 1\right) = 6$

$\log \left(\frac{2 x + 1}{x - 1}\right) = 6$

Since nothing is noted in subscript, the log is in base 10.

${10}^{6} = \left(\frac{2 x + 1}{x - 1}\right)$

$1 000 000 \left(x - 1\right) = 2 x + 1$

$1 000 000 x - 1 000 000 = 2 x + 1$

$999 998 x = 1 000 001$

$x = \left(\frac{1000001}{999 998}\right)$

Your teacher may want you to keep the answer in fractional form, but he/she may want it rounded off, so keep that in mind.

Practice exercises:

1. Solve for x. Leave answers in exact form.

a) ${\log}_{7} x = 3$

b) ${\log}_{3} \left(x + 1\right) = 5 - {\log}_{3} \left(4 {x}^{2} - 4\right)$

Good luck!