# How do you solve log (3x+1) = 1 + log (2x-1)?

Dec 23, 2015

The explanation is given below.

#### Explanation:

$\log \left(3 x + 1\right) = 1 + \log \left(2 x - 1\right)$
$\log \left(3 x + 1\right) = \log \left(10\right) + \log \left(2 x - 1\right)$ since $\log \left(10\right) = 1$
$\log \left(3 x + 1\right) = \log \left(10 \left(2 x - 1\right)\right)$ Product rule of logarithms
$3 x + 1 = 10 \left(2 x - 1\right)$ if logA = log B then A= B
$3 x + 1 = 20 x - 10$ distributive law

Now we have to solve for x using inverse operaitions to isolate x.

$3 x + 1 + 10 = 20 x - 10 + 10$

Adding 10 on both sides would remove $10$ from the right side of the equation

$3 x + 11 = 20 x$
$3 x + 11 - 3 x = 20 x - 3 x$

Subtracting $3 x$ on both sides would isolate the term containing $x$ to one side of the equation.

$11 = 17 x$

Dividing both sides by 17 we get

$\frac{11}{17} = \frac{17 x}{17}$

$\frac{11}{17} = x$

$x = \frac{11}{17}$ Answer