# How do you solve log ( x -3 ) + log ( x-5 ) = log ( 2x - 9) ?

Dec 17, 2015

$x = 12$

#### Explanation:

Given:

$\log \left(x - 3\right) + \log \left(x - 5\right) = \log \left(2 x - 9\right)$

Step 1: Rewrite the expression using sum to product rule

$\log \left[\left(x - 3\right) \left(x - 5\right)\right] = \log \left(2 x - 9\right)$

$\log \left({x}^{2} - 3 x - 5 x + 15\right) = \log \left(2 x - 9\right)$

Step 2 : Rewrite in exponential form with base to ("drop" log since we have sam log both side of equation)

${10}^{\log \left({x}^{2} - 8 x + 15\right)} = {10}^{\log \left(2 x - 9\right)}$

${x}^{2} - 8 x + 15 = 2 x - 9$

Step 3: Manipulate equation to write it in quadratic form $a {x}^{2} + b x + c = 0$

${x}^{2} - 8 x + 15 = 2 x - 9$

${x}^{2} - 10 x + 24 = 0$

Step 4: This can be solve by factoring

$\left(x - 12\right) \left(x + 2\right) = 0$ **

color(red)(x-12 = 0 => x= 12

$x + 2 = 0 \implies x = - 2$

Step 5: Check solution- can't have negative number as argument for the logarithm

Check $x = - 2$

$\log \left(- 2 - 3\right) + \log \left(- 2 - 5\right) = \log \left(2 \cdot - 2 - 9\right)$

$\log \left(- 5\right) + \log \left(- 7\right) = \log \left(- 13\right)$

Can't have negative as argument for logarithm, therefore $x = - 2$ is extraneous solution

** 2 number multiply equal to 24

like $- 12 \cdot 2 \mathmr{and} - 6 \cdot - 4 \mathmr{and} - 8 \cdot 3 \mathmr{and} - 2 \cdot 12 \text{ } e t c .$
**Add equal to $- 10$

$- 12 + 2 = - 10$