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

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Answer 1 · 2015-12-06T14:55:31

$b = 10 \Rightarrow x = \frac{3 + 7}{2} = 5$ $b = 10 Rightarrow x = (3 + 7)/2 = 5$

${log}_{b} [x (x - 3)] = {log}_{b} b$ $log_b [x(x-3)] = log_b b$

$x (x - 3) = b$ $x(x-3) = b$

$x^{2} - 3 x - b = 0$ $x^2 -3x - b = 0$

$x = \frac{3 \pm \sqrt{9 + 4 b}}{2}$ $x = (3 ± sqrt{9 + 4b})/2$

Consider $x > 0$ $x > 0$ and $x - 3 > 0$ $x - 3 > 0$

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