How do you solve the equation? 2log_3 (4x-5)= log_3(4x+5) + log_3(4x-3)

Nov 27, 2016

$x = \frac{5}{6}$

Explanation:

$\textcolor{b l u e}{2} {\log}_{3} \left(4 x - 5\right) = {\log}_{3} \left(4 x + 5\right) + {\log}_{3} \left(4 x - 3\right)$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times . x} \downarrow$
${\log}_{3} {\left(4 x - 5\right)}^{\textcolor{b l u e}{2}} = {\log}_{3} \left[\textcolor{red}{\left(4 x + 5\right) \times \left(4 x - 3\right)}\right]$

We have used using 2 log laws:

$\textcolor{b l u e}{{\log}_{a} {b}^{c} \Leftrightarrow c {\log}_{a} b} \mathmr{and} \textcolor{red}{{\log}_{a} b + {\log}_{a} c = {\log}_{a} \left(b \times c\right)}$

Now we can write:

${\left(4 x - 5\right)}^{2} = \left(4 x + 5\right) \times \left(4 x - 3\right)$

This uses another log law: ${\log}_{a} B = {\log}_{a} C \Leftrightarrow B = C$

Now solve it like a normal equation .... remove the brackets:

$16 {x}^{2} - 40 x + 25 = 16 {x}^{2} - 12 x + 20 x - 15$

$1 \cancel{6 {x}^{2}} - 40 x + 25 = \cancel{16 {x}^{2}} - 12 x + 20 x - 15$

$25 + 15 = 8 x + 40 x \text{ } \leftarrow$ re-arrange, keep $x$ positive.

$40 = 48 x$

$\frac{40}{48} = x$

$\frac{5}{6} = x$

Notice that, $\left(4 \left(\frac{5}{6}\right) - 5\right) < 0 \text{ } \leftarrow$ log impossible
However, the fact that it is squared gives a positive value, and the log value can then be determined.