How do you solve the equation ${log}_{3} (x + 5) - {log}_{3} (x - 7) = 2$ $log_3(x+5)-log_3(x-7)=2$ ?

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How do you solve the equation ${log}_{3} (x + 5) - {log}_{3} (x - 7) = 2$ $log_3(x+5)-log_3(x-7)=2$ ?

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Answer 1 · 2015-02-17T16:16:22

You can use the following facts:

${log}_{a} M - {log}_{a} N = log (\frac{M}{N})$ $log_aM-log_aN=log(M/N)$ and:

${log}_{a} b = x \Rightarrow a^{x} = b$ $log_ab=x => a^x=b$

So you get:

${log}_{3} (x + 5) - {log}_{3} (x - 7) = 2$ $log_3(x+5)-log_3(x-7)=2$
${log}_{3} (\frac{x + 5}{x - 7}) = 2$ $log_3((x+5)/(x-7))=2$
$\frac{x + 5}{x - 7} = 3^{2}$ $(x+5)/(x-7)=3^2$
$x + 5 = 9 (x - 7)$ $x+5=9(x-7)$
$x - 9 x = - 63 - 5$ $x-9x=-63-5$
$- 8 x = - 68$ $-8x=-68$
$x = \frac{68}{8} = \frac{17}{2}$ $x=68/8=17/2$

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How do you solve the equation ${log}_{3} (x + 5) - {log}_{3} (x - 7) = 2$ $log_3(x+5)-log_3(x-7)=2$ ?

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How do you solve the equation ${log}_{3} (x + 5) - {log}_{3} (x - 7) = 2$ $log_3(x+5)-log_3(x-7)=2$ ?