How do you solve $log (5 x + 2) = log (2 x - 5)$ $log(5x+2)=log(2x-5)$ ?

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How do you solve $log (5 x + 2) = log (2 x - 5)$ $log(5x+2)=log(2x-5)$ ?

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Answer 1 · 2015-12-09T06:22:03

$x = - \frac{7}{3}$ $x= -7/3$

Explanation:

Given $log (5 x + 2) = log (2 x - 5)$ $log(5x+2) = log(2x-5)$ common log- base 10

Step 1: Raised it to exponent using the base 10

$10^{log 5 x + 2} = 10^{log 2 x - 5}$ $10^(log5x+2) = 10^(log2x-5)$

Step 2: Simplify, since $10^{log A} = A$ $10^logA= A$
$5 x + 2 = 2 x - 5$ $5x+2= 2x-5$

Step 3: Subtract $2$ $color(red) 2$ and $2 x$ $color(blue)(2x)$ to both side of the equation to get
$5 x + 2 - 2 - 2 x = 2 x - 2 x - 5 - 2$ $5x+2color(red)(-2)color(blue)(-2x)= 2x color(blue)(-2x)-5color(red)(-2)$
$3 x = - 7$ $3x= -7$

Step 4: Dive both side by 3
$\frac{3 x}{3} = - \frac{7}{3} \Leftrightarrow x = - \frac{7}{3}$ $(3x)/3 = -7/3 hArr x= -7/3$

Step 5: Check the solution

$log [(5 \cdot - \frac{7}{3}) + 2] = log [(2 \cdot - \frac{7}{3}) - 5]$ $log[(5*-7/3)+2] = log[(2*-7/3) -5]$
$log (- \frac{35}{3} + \frac{6}{3}) = log (- \frac{14}{3} - \frac{15}{3})$ $log(-35/3 + 6/3) = log(-14/3 -15/3)$
$log (- \frac{29}{3}) = log (- \frac{29}{3})$ $log(-29/3) = log(-29/3)$

Both side are equal, despite we can't take a log of a negative number due to domain restriction ${log}_{b} x = y,, x > 0, b > 0$ $log_b x = y, , x >0 , b>0$
$x = - \frac{7}{3}$ $x= -7/3$ , assuming a complex valued logarithm

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How do you solve $log (5 x + 2) = log (2 x - 5)$ $log(5x+2)=log(2x-5)$ ?

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How do you solve log(5x+2)=log(2x−5)log(5x+2)=log(2x-5)?

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How do you solve $log (5 x + 2) = log (2 x - 5)$ $log(5x+2)=log(2x-5)$ ?