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

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Answer 1 · 2015-08-09T23:51:48

$color(red)(x=4/5)$

$log(5−2x)=log(3x+1)$

Convert the logarithmic equation to an exponential equation.

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

Remember that $10^logx =x$ , so

$5-2x=3x+1$

$4=5x$

$x=4/5$

Check:

$log(5−2x)=log(3x+1)$

If $x=4/5$ ,

$log(5−2(4/5))=log(3(4/5)+1)$

$log(5-8/5)=log(12/5+1)$

$log((25-8)/5)= log((12+5)/5)$

$log(17/5) = log(17/5)$

$x=4/5$ is a solution.

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