How do you solve log_[5] (24x + 4) - 2log_[5] (x - 2) = 2?

Apr 2, 2016

$x = 4$

Explanation:

$1$. Use the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, to rewrite $2 {\log}_{5} \left(x - 2\right)$.

${\log}_{5} \left(24 x + 4\right) - 2 {\log}_{5} \left(x - 2\right) = 2$

${\log}_{5} \left(24 x + 4\right) - {\log}_{5} \left({\left(x - 2\right)}^{2}\right) = 2$

$2$. Use the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\frac{\textcolor{red}{m}}{\textcolor{b l u e}{n}}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) - {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$ to simplify the left side of the equation.

${\log}_{5} \left(\frac{24 x + 4}{x - 2} ^ 2\right) = 2$

$3$. Use the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{p u r p \le}{b}}^{\textcolor{\mathmr{and} a n \ge}{x}}\right) = \textcolor{\mathmr{and} a n \ge}{x}$, to rewrite the right side of the equation.

${\log}_{5} \left(\frac{24 x + 4}{x - 2} ^ 2\right) = {\log}_{5} \left({5}^{2}\right)$

$4$. Since the equation now follows a "$\log = \log$" situation, where the bases are the same on both sides, rewrite the equation without the "$\log$" portion.

$\frac{24 x + 4}{x - 2} ^ 2 = {5}^{2}$

$5$. Isolate for $x$.

$24 x + 4 = 25 {\left(x - 2\right)}^{2}$

$24 x + 4 = 25 \left({x}^{2} - 4 x + 4\right)$

$24 x + 4 = 25 {x}^{2} - 100 x + 100$

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{25} {x}^{2}$ $\textcolor{t u r q u o i s e}{- 124} x$ $\textcolor{v i o \le t}{+ 96} = 0$

$6$. Use the quadratic formula to solve for $x$.

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{a = 25} \textcolor{w h i t e}{X X X X X} \textcolor{t u r q u o i s e}{b = - 124} \textcolor{w h i t e}{X X X X X} \textcolor{v i o \le t}{c = 96}$

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

$x = \frac{- \left(\textcolor{t u r q u o i s e}{- 124}\right) \pm \sqrt{{\left(\textcolor{t u r q u o i s e}{- 124}\right)}^{2} - 4 \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{25}\right) \left(\textcolor{v i o \le t}{96}\right)}}{2 \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{25}\right)}$

$x = \frac{124 \pm \sqrt{15376 - 9600}}{50}$

$x = \frac{124 \pm \sqrt{5776}}{50}$

$x = \frac{124 \pm 76}{50}$

$x = \frac{124 + 76}{50} \textcolor{w h i t e}{i} , \textcolor{w h i t e}{i} \frac{124 - 76}{50}$

$x = \frac{200}{50} \textcolor{w h i t e}{i} , \textcolor{w h i t e}{i} \frac{48}{50}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = 4 \textcolor{w h i t e}{i} , \textcolor{w h i t e}{i} \textcolor{red}{\cancel{\textcolor{g r e e n}{\frac{24}{25}}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

However, if you substitute $x = \frac{24}{25}$ back into the original equation, you will find that you will end up taking the logarithm of a $\textcolor{red}{\text{negative number}}$, which is $\textcolor{red}{\text{not possible}}$.

For example:

${\log}_{5} \left(24 x + 4\right) - 2 {\log}_{5} \left(x - 2\right) = 2$

${\log}_{5} \left(24 \left(\frac{24}{25}\right) + 4\right) - 2 {\log}_{5} \left(\frac{24}{25} - 2\right) = 2$

${\log}_{5} \left(\frac{676}{25}\right) - 2 {\log}_{5} \left(\textcolor{red}{- \frac{26}{25}}\right) = 2$

For this reason, the correct and only solution to the given logarithmic equation is $x = 4$.