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How do you solve $ln (y - 1) - ln (2) = x + ln x$ $ln(y-1) - ln(2) = x + ln x$ ?

1 Answer

Dec 12, 2015

Use property of logs ...

Explanation:

$ln [\frac{y - 1}{2 \times x}] = x$ $ln[(y-1)/(2xxx)]=x$

Exponentiate ...

$e^{ln [\frac{y - 1}{2 \times x}]} = e^{x}$ $e^{ln[(y-1)/(2xxx)]}=e^x$

Simplify ...

$\frac{y - 1}{2 x} = e^{x}$ $(y-1)/(2x)=e^x$

$y = (2 x) e^{x} + 1$ $y=(2x)e^x+1$

This function approaches $y = 1$ $y=1$ as $x \to - \infty$ $xrarr-oo$ . It never crosses the x-axis.

Hope that helps

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