How do you simplify log (x+6)=1-log(x-5)?

May 11, 2018

I tried this:

Explanation:

We can use, as an example, the base of the logs as being $10$. We get:

${\log}_{10} \left(x + 6\right) = 1 - {\log}_{10} \left(x - 5\right)$

rearrange:

${\log}_{10} \left(x + 6\right) + {\log}_{10} \left(x - 5\right) = 1$

use a property of logs to write:

${\log}_{10} \left[\left(x + 6\right) \left(x - 5\right)\right] = 1$

use the definition of log to write:

$\left(x + 6\right) \left(x - 5\right) = {10}^{1}$

that becomes:

${x}^{2} - 5 x + 6 x - 30 - 10 = 0$
${x}^{2} + x - 40 = 0$

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{1 + 160}}{2} = \frac{- 1 \pm \sqrt{161}}{2}$
and:
${x}_{1} = \frac{- 1 + \sqrt{161}}{2} = 5.8442$
${x}_{2} = \frac{- 1 - \sqrt{161}}{2} = - 6.8442$

the negative solution cannot be accepted.

May 11, 2018

$x = \frac{- 1 + \sqrt{161}}{2} \approx 5.844$

Explanation:

$\log \left(x + 6\right) = 1 - \log \left(x - 5\right)$

Move $\log \left(x - 5\right)$ to Left Side
$\log \left(x + 6\right) + \log \left(x - 5\right) = 1$

Condense the Left Side into 1 Log expression
$\log \left(\left(x + 6\right) \left(x - 5\right)\right) = 1$

Convert from Log Equation to Exponential Equation
$\left(x + 6\right) \left(x - 5\right) = {10}^{1}$

Multiply Left Side
${x}^{2} - x - 30 = 10$

Move $10$ to Left Side
${x}^{2} - x - 40 = 0$

Since we can't factor, use Quadratic Formula:
Quadratic Formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 1 + \sqrt{161}}{2} \approx 5.844$
And
$x = \frac{- 1 - \sqrt{161}}{2} \approx - 6.844$

Since the "inside" of log has to be positive,
$x = \frac{- 1 + \sqrt{161}}{2} \approx 5.844$ is the answer.