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

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

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Answer 1 · 2016-01-06T18:45:44

$x = 5$ $x=5$

Explanation:

$log x + log (x - 3) = 1$ $logx+log(x-3)=1$
We know that: $log a + log b = log (a \cdot b)$ $loga+logb=log(a*b)$
$\Rightarrow log (x (x - 3)) = 1$ $implies Log(x(x-3))=1$
$\Rightarrow log (x^{2} - 3 x) = 1$ $implies log(x^2-3x)=1$
$\Rightarrow x^{2} - 3 x = 10$ $implies x^2-3x=10$
$\Rightarrow x^{2} - 3 x - 10 = 0$ $implies x^2-3x-10=0$
$\Rightarrow (x - 5) (x + 2) = 0$ $implies (x-5)(x+2)=0$
$\Rightarrow x = 5, - 2$ $implies x=5,-2$

Verification:-
Put $x = 5$ $x=5$
$L . H . S = log x + log (x - 3) = log 5 + log (5 - 3) = log 5 + log 2 = log (5 \cdot 2) = log 10 = 1 = R . H . S$ $L.H.S=Logx+Log(x-3)=Log5+log(5-3)=log5+log2=log(5*2)=log10=1=R.H.S$
Verified.
Put $x = - 2$ $x=-2$
$L . H . S = log x + log (x - 3) = log (- 2) + log (- 2 - 3) = log (- 2) + log (- 5)$ $L.H.S=Logx+Log(x-3)=Log(-2)+log(-2-3)=log(-2)+log(-5)$
Here we have to find the log of a negative number which is undefined.
Therefore not verified.

Therefore, only $x = 5$ $x=5$ is true.

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

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