How do you solve $ln x + ln (x + 5) = ln 14$ $ln x + ln (x+5) = ln 14$ ?

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How do you solve $ln x + ln (x + 5) = ln 14$ $ln x + ln (x+5) = ln 14$ ?

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Answer 1 · 2015-12-19T21:35:32

I found $x = 2$ $x=2$

Explanation:

Use the property of the log:
$log x + log y = log (x \cdot y)$ $logx+logy=log(x*y)$
to get:
$ln [x (x + 5)] = ln (14)$ $ln[x(x+5)]=ln(14)$
for the logs to be equal the arguments must be equal as well, or:
$x (x + 5) = 14$ $x(x+5)=14$
solve for $x$ $x$ :
$x^{2} + 5 x - 14 = 0$ $x^2+5x-14=0$
Use the Quadratic Formula:
$x_{1, 2} = \frac{- 5 \pm \sqrt{25 + 56}}{2} = \frac{- 5 \pm \sqrt{81}}{2} = \frac{- 5 \pm 9}{2} =$ $x_(1,2)=(-5+-sqrt(25+56))/2=(-5+-sqrt(81))/2=(-5+-9)/2=$
two solutions:
$x_{1} = - 7$ $x_1=-7$ NO it gives you a negative argument in the original logs.
$x_{2} = 2$ $x_2=2$ YES

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How do you solve $ln x + ln (x + 5) = ln 14$ $ln x + ln (x+5) = ln 14$ ?

1 Answer

Explanation:

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