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

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

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Answer 1 · 2016-03-11T14:53:44

$\Rightarrow x = - 5 and x = 2$ $=>x=-5 and x=2$

Explanation:

$log (x + 3) + log x = 1$ $log(x+3)+logx =1$
$\Rightarrow log ((x + 3) \times x) = log 10$ $=>log((x+3)xxx) =log10$
$\Rightarrow x^{2} + 3 x = 10$ $=>x^2+3x=10$
$\Rightarrow x^{2} + 3 x - 10 = 0$ $=>x^2+3x-10=0$
$\Rightarrow x^{2} + 5 x - 2 x - 10 = 0$ $=>x^2+5x-2x-10=0$
$\Rightarrow x (x + 5) - 2 (x + 5) = 0$ $=>x(x+5)-2(x+5)=0$
$\Rightarrow (x + 5) (x - 2) = 0$ $=>(x+5)(x-2)=0$
$\Rightarrow x = - 5 and x = 2$ $=>x=-5 and x=2$

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

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