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

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

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Answer 1 · 2015-12-03T14:22:07

$\Rightarrow x_{1} = - 2, x_{2} = 5$ $=>x_1=-2, x_2=5$

Explanation:

$log (x + 3) = 1 - log (x)$ $log(x+3)=1-log(x)$
$\Rightarrow log (x + 3) + log (x) = 1 - log (x) + log (x)$ $=>log(x+3)+log(x)=1-log(x)+log(x)$
The logarithm of a product is the sum of the logarithms of the numbers being multiplied:
$log (x^{2} - 3 x) = 1$ $log(x^2-3x)=1$
$\Rightarrow x^{2} - 3 x = 10$ $=>x^2-3x=10$
$\Rightarrow x^{2} - 3 x - 10 = 10 - 10$ $=>x^2-3x-10=10-10$
$\Rightarrow x^{2} - 3 x - 10 = 0$ $=>x^2-3x-10=0$
$\Rightarrow (x - 5) (x + 2) = 0$ $=>(x-5)(x+2)=0$
$\Rightarrow x_{1} = - 2, x_{2} = 5$ $=>x_1=-2, x_2=5$

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

1 Answer

Explanation:

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