How do you solve $log 3 x = log 2 + log (x + 5)$ $log3x=log2+log (x+5)$ ?

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How do you solve $log 3 x = log 2 + log (x + 5)$ $log3x=log2+log (x+5)$ ?

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Answer 1 · 2016-04-08T10:58:51

$x = 10$ $x = 10$

Explanation:

Rearrange slightly to get $x$ $x$ on one side and constant on the other

$log 3 x = log 2 + log (x + 5)$ $log3x = log 2 + log(x + 5)$
$log 3 + log x = log 2 + log (x + 5)$ $log3 + logx = log 2 + log(x+5)$
$log x - log (x + 5) = log 2 - log 3$ $logx - log(x+5) = log2 - log3$

Now, using laws of logarithms, make it so you only have a single logarithm on either side

$log x - log (x + 5) = log (\frac{x}{x + 5})$ $logx - log(x+5) = log(x/(x+5))$
$log 2 - log 3 = log (\frac{2}{3})$ $log2 - log3 = log(2/3)$

So you have

$log (\frac{x}{x + 5}) = log (\frac{2}{3})$ $log(x/(x+5)) = log(2/3)$

Raise both sides to the power $e$ $e$ to cancel out the logarithms (assuming we are dealing with ${log}_{e}$ $log_e$ or $ln$ $ln$ ),

$\frac{x}{x + 5} = \frac{2}{3}$ $x/(x+5) = 2/3$

From which we get

$\frac{x}{x + 5} = \frac{2}{3}$ $x/(x+5) = 2/3$
$3 x = 2 (x + 5)$ $3x = 2(x+5)$
$3 x = 2 x + 10$ $3x = 2x + 10$
$3 x - 2 x = 10$ $3x - 2x = 10$
$x = 10$ $x = 10$

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How do you solve $log 3 x = log 2 + log (x + 5)$ $log3x=log2+log (x+5)$ ?

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How do you solve log3x=log2+log(x+5)log3x=log2+log (x+5)?

1 Answer

Explanation:

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How do you solve $log 3 x = log 2 + log (x + 5)$ $log3x=log2+log (x+5)$ ?