How do you solve ${ln e}^{ln x} - ln (x - 2) = ln 8$ $ln e^ln x - ln(x-2) = ln 8$ ?

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Answer 1 · 2016-06-18T02:24:04

$x = \frac{16}{7}$ $x=16/7$

${ln e}^{ln x} - ln (x - 2) = ln 8$ $ln e^ln x - ln(x-2) = ln 8$

$\Rightarrow ln x - ln (x - 2) = ln 8$ $=>ln x - ln(x-2) = ln 8$

$\Rightarrow ln (\frac{x}{x - 2}) = ln 8$ $=>ln (x/(x-2) )= ln 8$

$\Rightarrow (\frac{x}{x - 2}) = 8$ $=> (x/(x-2) )= 8$

$\Rightarrow 8 x - 16 = x$ $=>8x-16=x$

$\Rightarrow 7 x = 16 \Rightarrow x = \frac{16}{7}$ $=>7x=16=>x=16/7$

How do you solve lnelnx−ln(x−2)=ln8ln e^ln x - ln(x-2) = ln 8?