How do you solve $log x + log (x + 15) = 2$ $logx + log(x + 15) = 2$ ?

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How do you solve $log x + log (x + 15) = 2$ $logx + log(x + 15) = 2$ ?

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Answer 1 · 2016-03-21T07:05:54

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Explanation:

$log x + log (x + 15) = 2$ $logx+log(x+15)=2$
$log (x (x + 15)) = 2$ $log(x(x+15))=2$

Here is the catch. This log can be from any base p. I assume this is log base 10. So
${log}_{10} (x^{2} + 15 x) = 2$ $log_10(x^2+15x)=2$
Taking antilog on both sides
$x^{2} + 15 x = 10^{2}$ $x^2+15x=10^2$
$x^{2} + 15 x - 100 = 0$ $x^2+15x-100=0$

Then factorize this to get the solution.

$(x + 20) \times (x - 5) = 0$ $(x+20)\times(x-5) = 0$
$x = - 20, 5$ $x = -20,5$

Similar equations can be derived for various bases to get different solutions.

For general case the equation to solve is
$x^{2} + 15 x - p^{2} = 0$ $x^2+15x-p^2=0$
$x = \frac{- 15 \pm \sqrt{225 + 4 p^{2}}}{2}$ $x = (-15\pmsqrt(225+4p^2))/2$

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How do you solve $log x + log (x + 15) = 2$ $logx + log(x + 15) = 2$ ?

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Science

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Humanities

... and beyond

How do you solve logx+log(x+15)=2 logx + log(x + 15) = 2?

1 Answer

Explanation:

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Impact of this question

How do you solve $log x + log (x + 15) = 2$ $logx + log(x + 15) = 2$ ?