How do you solve for x in log_ 6 (2-x) + log _ 6 (3-4x) = 1?

Feb 6, 2016

$x = 0$

Explanation:

Property of Logarithmic expression

$\log A + \log B = L o g \left(A B\right) \text{ " " " } \left(1\right)$
$n \log A = \log {A}^{n} \text{ " " } \left(2\right)$
${\log}_{a} y = x \implies {a}^{x} = y \text{ " } \left(3\right)$

Given :

${\log}_{6} \left(2 - x\right) + {\log}_{6} \left(3 - 4 x\right) = 1$

Rewrite as:

Using rule (2)

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

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

${\log}_{6} \left(6 - 5 x + {x}^{2}\right) = 1$

Using rule (3)

${6}^{1} = 6 - 5 {x}^{2} + {x}^{2}$

Now we have a simple quadratic equation, let's solve it.

${x}^{2} - 5 x + 6 = 6$

$- 6 \text{ " } - 6$

$= = = = = = = =$

${x}^{2} - 5 x = 0$
$x \left(x - 5\right) = 0$

$x = 0$ or $x = 5$

Always check your answer if you have 2 solutions when solving a log equation .

$\textcolor{b l u e}{C h e c k \text{ } x = 5}$

$x = 5$

${\log}_{6} \left(2 - 5\right) + {\log}_{6} \left(3 - 4 \left(5\right)\right) = 1$

${\log}_{6} \left(- 3\right) + {\log}_{6} \left(- 17\right) = 1$

$\textcolor{red}{x = 5 \text{ Extraneous } s o l u t i o n}$

Can't evaluate the expression on the left, because the argument of any logarithm always POSITIVE and greater than zero, due to domain restriction

$\textcolor{b l u e}{C h e c k \text{ } x = 0}$

color(red)(log_6(2-0)+log_6(3-4(0))= 1

${\log}_{6} \left(2\right) + {\log}_{6} \left(3\right) = 1$

${\log}_{6} \left(6\right) = 1$

$1 = 1$

Solution $\textcolor{red}{x = 0}$