How do you solve ${log}_{5} (24 x + 4) - 2 {log}_{5} (x - 2) = 2$ $log_[5] (24x + 4) - 2log_[5] (x - 2) = 2$ ?

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How do you solve ${log}_{5} (24 x + 4) - 2 {log}_{5} (x - 2) = 2$ $log_[5] (24x + 4) - 2log_[5] (x - 2) = 2$ ?

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Answer 1 · 2016-04-02T02:30:16

$x = 4$ $x=4$

Explanation:

$1$ $1$ . Use the logarithmic property, ${log}_{b} (m^{n}) = n \cdot {log}_{b} (m)$ $log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)$ , to rewrite $2 {log}_{5} (x - 2)$ $2log_5(x-2)$ .

${log}_{5} (24 x + 4) - 2 {log}_{5} (x - 2) = 2$ $log_5(24x+4)-2log_5(x-2)=2$

${log}_{5} (24 x + 4) - {log}_{5} ({(x - 2)}^{2}) = 2$ $log_5(24x+4)-log_5((x-2)^2)=2$

$2$ $2$ . Use the logarithmic property, ${log}_{b} (\frac{m}{n}) = {log}_{b} (m) - {log}_{b} (n)$ $log_color(purple)b(color(red)m/color(blue)n)=log_color(purple)b(color(red)m)-log_color(purple)b(color(blue)n)$ to simplify the left side of the equation.

${log}_{5} (\frac{24 x + 4}{{(x - 2)}^{2}}) = 2$ $log_5((24x+4)/(x-2)^2)=2$

$3$ $3$ . Use the logarithmic property, ${log}_{b} (b^{x}) = x$ $log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x$ , to rewrite the right side of the equation.

${log}_{5} (\frac{24 x + 4}{{(x - 2)}^{2}}) = {log}_{5} (5^{2})$ $log_5((24x+4)/(x-2)^2)=log_5(5^2)$

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

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

$5$ $5$ . Isolate for $x$ $x$ .

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

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

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

$25 x^{2}$ $color(darkorange)25x^2$ $- 124 x$ $color(turquoise)(-124)x$ $+ 96 = 0$ $color(violet)(+96)=0$

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

$a = 25 X X X X X b = - 124 X X X X X c = 96$ $color(darkorange)(a=25)color(white)(XXXXX)color(turquoise)(b=-124)color(white)(XXXXX)color(violet)(c=96)$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- (- 124) \pm \sqrt{{(- 124)}^{2} - 4 (25) (96)}}{2 (25)}$ $x=(-(color(turquoise)(-124))+-sqrt((color(turquoise)(-124))^2-4(color(darkorange)(25))(color(violet)(96))))/(2(color(darkorange)(25)))$

$x = \frac{124 \pm \sqrt{15376 - 9600}}{50}$ $x=(124+-sqrt(15376-9600))/50$

$x = \frac{124 \pm \sqrt{5776}}{50}$ $x=(124+-sqrt(5776))/50$

$x = \frac{124 \pm 76}{50}$ $x=(124+-76)/50$

$x = \frac{124 + 76}{50} i, i \frac{124 - 76}{50}$ $x=(124+76)/50color(white)(i),color(white)(i)(124-76)/50$

$x = \frac{200}{50} i, i \frac{48}{50}$ $x=200/50color(white)(i),color(white)(i)48/50$

$color(green)(|bar(ul(color(white)(a/a)x=4color(white)(i),color(white)(i)color(red)cancelcolor(green)(24/25)color(white)(a/a)|)))$

However, if you substitute $x=24/25$ back into the original equation, you will find that you will end up taking the logarithm of a $color(red)("negative number")$ , which is $color(red)("not possible")$ .

For example:

$log_5(24x+4)-2log_5(x-2)=2$

$log_5(24(24/25)+4)-2log_5(24/25-2)=2$

$log_5(676/25)-2log_5(color(red)(-26/25))=2$

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

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How do you solve ${log}_{5} (24 x + 4) - 2 {log}_{5} (x - 2) = 2$ $log_[5] (24x + 4) - 2log_[5] (x - 2) = 2$ ?

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Explanation:

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How do you solve log_[5] (24x + 4) - 2log_[5] (x - 2) = 2log5(24x+4)−2log5(x−2)=2log_[5] (24x + 4) - 2log_[5] (x - 2) = 2?

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How do you solve ${log}_{5} (24 x + 4) - 2 {log}_{5} (x - 2) = 2$ $log_[5] (24x + 4) - 2log_[5] (x - 2) = 2$ ?