# How do you solve log (x + 3) = log (6) - log (2x-1)?

Nov 25, 2015

$x = \frac{- 5 + \sqrt{97}}{4}$

#### Explanation:

I'm assuming that all the $\log$ functions have the same basis $> 1$.

First of all, let's compute the domain of the logarithmic expressions.
For the first one, $x + 3 > 0$ must hold, so $x > - 3$.
For the last one, we get $2 x - 1 > 0 \iff x > \frac{1}{2}$.

As the second condition is the more restrictive one, we can assume that the domain is $x > \frac{1}{2}$ and any possible solutions need to respect this condition.

Now, in order to "get rid" of the logarithmic expressions, it is necessary to simplify the terms on the right-hand side.

To do so, remember the logarithmic rule: ${\log}_{a} \left(x\right) - {\log}_{a} \left(y\right) = {\log}_{a} \left(\frac{x}{y}\right)$

In our case, it means:

$\log \left(x + 3\right) = \log \left(6\right) - \log \left(2 x - 1\right)$
$\iff \log \left(x + 3\right) = \log \left(\frac{6}{2 x - 1}\right)$

Now, due to the fact that for $x$, $y > 0$ and $a \ne 1$, $\log x = \log y \iff x = y$ holds, we can "drop" the logarithms on both sides of the equation.

$\iff x + 3 = \frac{6}{2 x - 1}$

... multiply both sides with $2 x - 1$...

$\iff \left(x + 3\right) \left(2 x - 1\right) = 6$

$\iff 2 {x}^{2} + 5 x - 9 = 0$

This is a quadratic equation that can be solved e. g. with the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Here, $a = 2$, $b = 5$, $c = - 9$, so we can compute the solution as follows:

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 2 \cdot \left(- 9\right)}}{4} = \frac{- 5 \pm \sqrt{97}}{4}$

The solution $x = \frac{- 5 - \sqrt{97}}{4}$ is negative, therefore we can ignore it because of our restriction to the domain $x > \frac{1}{2}$.

So, we have just one solution: $x = \frac{- 5 + \sqrt{97}}{4} \approx 1.2122$

Hope that this helped!