How do you solve $log (5 - 2 x) = log (3 x + 1)$ $log(5-2x)=log(3x +1)$ ?

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Answer 1 · 2015-08-09T23:51:48

$x = \frac{4}{5}$ $color(red)(x=4/5)$

$log (5 - 2 x) = log (3 x + 1)$ $log(5−2x)=log(3x+1)$

Convert the logarithmic equation to an exponential equation.

$10^{log (5 - 2 x)} = 10^{log (3 x + 1)}$ $10^(log(5-2x)) = 10^(log(3x+1))$

Remember that $10^{log x} = x$ $10^logx =x$ , so

$5 - 2 x = 3 x + 1$ $5-2x=3x+1$

$4 = 5 x$ $4=5x$

$x = \frac{4}{5}$ $x=4/5$

Check:

$log (5 - 2 x) = log (3 x + 1)$ $log(5−2x)=log(3x+1)$

If $x = \frac{4}{5}$ $x=4/5$ ,

$log (5 - 2 (\frac{4}{5})) = log (3 (\frac{4}{5}) + 1)$ $log(5−2(4/5))=log(3(4/5)+1)$

$log (5 - \frac{8}{5}) = log (\frac{12}{5} + 1)$ $log(5-8/5)=log(12/5+1)$

$log (\frac{25 - 8}{5}) = log (\frac{12 + 5}{5})$ $log((25-8)/5)= log((12+5)/5)$

$log (\frac{17}{5}) = log (\frac{17}{5})$ $log(17/5) = log(17/5)$

$x = \frac{4}{5}$ $x=4/5$ is a solution.

How do you solve log(5-2x)=log(3x +1)log(5−2x)=log(3x+1)log(5-2x)=log(3x +1)?