How do you solve $log (x + 7) - log (x - 7) = 1$ $log(x+7) - log(x-7)=1$ ?

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Answer 1 · 2016-07-16T07:21:09

$x = \frac{77}{9}$ $x=77/9$

As $log (x + 7) - log (x - 7) = 1$ $log(x+7)−log(x−7)=1$ , we have

$log (\frac{x + 7}{x - 7}) = 1 = log 10$ $log((x+7)/(x−7))=1 = log 10$

$\frac{x + 7}{x - 7} = 10$ $(x+7)/(x−7)=10$ or

$x + 7 = 10 x - 70$ $x+7=10x-70$ or

$10 x - x = 77$ $10x-x=77$ or

$9 x = 77$ $9x=77$ or

$x = \frac{77}{9}$ $x=77/9$

How do you solve log(x+7)−log(x−7)=1log(x+7) - log(x-7)=1?