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

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Answer 1 · 2015-12-06T23:32:20

$x=5$

Rearrange the expression to get

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

Now, use the property $log(a)+log(b)=log(a*b)$ :

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

If by $log(x)$ you mean the logarithm in base $10$ , you can write $1$ as $log(10)$ , and so the expression becomes

$log(x(x-3))=log(10)$

Now use the fact that $log(a)=log(b) \iff a=b$ :

$x(x-3)=10$

Expand:

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

To solve this equation, we need two numbers $x_1$ and $x_2$ such that:

$x_1+x_2=3$
$x_1*x_2=-10$

These numbers are $5$ and $-2$ . We can't accept $-2$ as a solution, because it would lead to the logarithm of a negative number.

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