How do you solve ${log}_{4} (x + 1) - {log}_{4} (x - 2) = 3$ $log_4(x+1) - log_4(x-2) = 3$ ?

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How do you solve ${log}_{4} (x + 1) - {log}_{4} (x - 2) = 3$ $log_4(x+1) - log_4(x-2) = 3$ ?

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Answer 1 · 2015-12-06T16:55:22

I found: $x = 2.04762$ $x=2.04762$

Explanation:

You can use the property of logs that tells us:
$log x - log y = log (\frac{x}{y})$ $logx-logy=log(x/y)$
to get:
${log}_{4} (\frac{x + 1}{x - 2}) = 3$ $log_4((x+1)/(x-2))=3$
Now we use the definition of log to change it into an exponential:
$\frac{x + 1}{x - 2} = 4^{3}$ $(x+1)/(x-2)=4^3$
$\frac{x + 1}{x - 2} = 64$ $(x+1)/(x-2)=64$
$(x + 1) = 64 (x - 2)$ $(x+1)=64(x-2)$
$x + 1 = 64 x - 128$ $x+1=64x-128$
$63 x = 129$ $63x=129$
$x = \frac{129}{63} = 2.04762$ $x=129/63=2.04762$

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How do you solve ${log}_{4} (x + 1) - {log}_{4} (x - 2) = 3$ $log_4(x+1) - log_4(x-2) = 3$ ?

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How do you solve ${log}_{4} (x + 1) - {log}_{4} (x - 2) = 3$ $log_4(x+1) - log_4(x-2) = 3$ ?