How do you solve $log (x + 9) = log (2 x - 7)$ $log (x+9) = log (2x-7)$ ?

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Answer 1 · 2015-08-01T19:14:26

$x = 16$ $color(red)(x=16)$

$log (x + 9) = log (2 x - 7)$ $log(x+9) = log (2x-7)$

Convert the logarithmic equation to an exponential equation.

$10^{log (x + 9)} = 10^{log (2 x - 7)}$ $10^( log(x+9)) = 10^( log (2x-7))$

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

$x + 9 = 2 x - 7$ $x+9 = 2x-7$

Move all terms to the right hand side.

$0 = 2 x - 7 - x - 9$ $0 =2x-7-x-9$

Combine like terms.

$0 = x - 16$ $0 =x-16$

$x = 16$ $x=16$

Check:

$log (x + 9) = log (2 x - 7)$ $log(x+9) = log (2x-7)$

If $x = 16$ $x=16$

$log (16 + 9) = log (2 (16) - 7)$ $log(16+9) = log (2(16)-7)$

$log 25 = log (32 - 7)$ $log25 = log (32-7)$

$log 25 = log 25$ $log25 = log25$

$x = 16$ $x=16$ is a solution.

How do you solve log(x+9)=log(2x−7)log (x+9) = log (2x-7)?