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

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Answer 1 · 2018-03-03T18:49:39

The solution is $x=2$ .

$log_color(green)a(color(red)x)+log_color(green)a(color(blue)y)=log_color(green)a(color(red)xcolor(blue)y)$

Now combine the two $log$ s, then rewrite $1$ as $log(10)$ , then cancel the logs on either side:

$log(x)+log(x+3)=1$

$log(x*(x+3))=1$

$log(x^2+3x)=1$

$log(x^2+3x)=log(10)$

$color(red)cancel(color(black)log)(x^2+3x)=color(red)cancel(color(black)log)(10)$

$x^2+3x=10$

$x^2+3x-10=0$

$(x+5)(x-2)=0$

$x=2,-5$

Plug in the answers to the original equation and see if they still work:

$log(x)+log(x+3)=1$

Testing $2$ :

$log(2)+log(2+3)=1$

$log(2)+log(5)=1$

$0.30102...+0.69898...=1$

$1=1qquadqquad color(lightgreen)sqrt$

The solution $2$ works. Testing $-5$ :

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

Since $log(-5)$ is undefined, this solution doesn't work. Therefore, the only solution is $x=2$ .

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