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

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Answer 1 · 2015-11-21T15:01:13

$x=13$

To do this problem, you must know that $log(a)+log(b)=log(ab)$ and that $log(a)-log(b)=log(a/b)$ .

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

$log(x+5)+log(x-3)-log(x-1)-log(x+2)=0$

$log(((x+5)(x-3))/((x-1)(x+2)))=0$

Remember that $log(a)$ is another way of writing $log_10(a)$ .

$10^(log(((x+5)(x-3))/((x-1)(x+2))))=10^0$

$((x+5)(x-3))/((x-1)(x+2))=1$

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

$x^2+2x-15=x^2+x-2$

$2x-15=x-2$

$x=13$

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