How do you solve $log x + log (x - 9) = 1$ $Log x + log(x-9)=1$ ?

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How do you solve $log x + log (x - 9) = 1$ $Log x + log(x-9)=1$ ?

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Answer 1 · 2015-08-10T00:15:08

$x = 10$ $color(red)(x=10)$

Explanation:

$log x + log (x - 9) = 1$ $logx+log(x−9)=1$

Recall that $log a + log b = log (a b)$ $loga+logb=log(ab)$ , so

$log x + log (x - 9) = log (x (x - 9)) = log (x^{2} - 9 x)$ $logx+log(x−9)=log(x(x-9))=log(x^2-9x)$

$log (x^{2} - 9 x) = 1$ $log(x^2-9x)=1$

Convert the logarithmic equation to an exponential equation.

$10^{log (x^{2} - 9 x)} = 10^{1}$ $10^(log(x^2-9x)) = 10^1$

Remember that $10^{log x} = x$ $10^logx =x$ , so

$x^{2} - 9 x = 10$ $x^2-9x=10$

$x^{2} - 9 x - 10 = 0$ $x^2-9x-10=0$

$(x - 10) (x + 1) = 0$ $(x-10)(x+1)=0$

$x - 10 = 0$ $x-10=0$ and $x + 1 = 0$ $x+1=0$

$x = 10$ $x=10$ and $x = - 1$ $x=-1$

Check:

$log x + log (x - 9) = 1$ $logx+log(x−9)=1$

If $x = 10$ $x=10$ ,

$log 10 + log (10 - 9) = 1$ $log10+log(10−9)=1$

$log 10 + log 1 = 1$ $log10+log1=1$

$1 + 0 = 1$ $1+0=1$

$1 = 1$ $1=1$

$x = 10$ $x=10$ is a solution.

If $x = - 1$ $x=-1$ ,

$log (- 1) + log (- 1 - 9) = 1$ $log(-1)+log(-1-9)=1$

This is impossible, because the logarithm of a negative number is undefined.

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How do you solve $log x + log (x - 9) = 1$ $Log x + log(x-9)=1$ ?

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How do you solve logx+log(x−9)=1Log x + log(x-9)=1?

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How do you solve $log x + log (x - 9) = 1$ $Log x + log(x-9)=1$ ?