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

Question

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

1 Answer

Impact of this question

2655 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-03T18:23:39

$x = 12$ $x= 12$

Explanation:

Re-write as single logarithmic expression

Note: $log (a) - log (b) = log (\frac{a}{b})$ $log(a) - log(b) = log(a/b)$

$log (2 + x) - log (x - 5) = log 2$ $log(2+x) - log(x-5) = log2$

$log (\frac{2 + x}{x - 5}) = log 2$ $log((2+x)/(x-5))= log 2$

$10^{log (\frac{2 + x}{x - 5})} = 10^{log 2}$ $10^log((2+x)/(x-5)) = 10^(log2)$

$\frac{2 + x}{x - 5} = 2$ $(2+x)/(x-5) = 2$

$\frac{2 + x}{x - 5} \cdot (x - 5) = 2 \cdot (x - 5)$ $(2+x)/(x-5)*color(red)((x-5)) = 2*color(red)((x-5))$

$(2+x)/cancel(x-5)*cancel((x-5)) = 2(x-5)$

$2+x " " "= 2x- 10$
$+10 - x = -x +10$

===============
$color(red)(12 " " " = x)$

Check :
$log(12+2) - log(12-5) = log 2$ ?
$log(14) - log(7)$
$log(14/7)$
$log 2 = log 2$

Yes, answer is $x= 12$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve log(2+x)-log(x-5)=log 2log(2+x)−log(x−5)=log2log(2+x)-log(x-5)=log 2?

1 Answer

Explanation:

Related questions

Impact of this question

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