How do you solve $ln x - ln (\frac{1}{x}) = 2$ $ln x - ln (1/x) = 2$ ?

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How do you solve $ln x - ln (\frac{1}{x}) = 2$ $ln x - ln (1/x) = 2$ ?

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Answer 1 · 2018-03-16T19:37:06

$x = e$ $x=e$

Explanation:

Since $ln (\frac{1}{x}) = - ln x$ $ln(1/x) = - ln x$ , the equation becomes

$2 ln x = 2 \Rightarrow ln x = 1$ $2 ln x = 2 implies ln x =1$

So

$x = e$ $x=e$

Answer 2 · 2018-03-16T19:37:47

$x = e$ $x=e$

Explanation:

We have, $(1) {log}_{a} X = n \Leftrightarrow X = a^{n}$ $(1) log_aX=n<=>X=a^n$
$(2) {log}_{a} (\frac{M}{N}) = {log}_{a} M - {log}_{a} N$ $(2)log_a(M/N)=log_aM-log_aN$
Here,
$ln x - ln (\frac{1}{x}) = 2$ $lnx-ln(1/x)=2$ , Applying (2) ,we get
$\Rightarrow ln x - [ln 1 - ln x] = 2$ $=>lnx-[ln1-lnx]=2$
$\Rightarrow ln x - ln 1 + ln x = 2$ $=>lnx-ln1+lnx=2$ , where, $ln 1 = 0$ $ln1=0$
$\Rightarrow 2 ln x = 2$ $=>2lnx=2$
$\Rightarrow ln x = 1 \Rightarrow {log}_{e} x = 1$ $=>lnx=1=>log_ex=1$ , Applying (1) ,we get
$\Rightarrow x = e^{1} = e$ $=>x=e^1=e$

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How do you solve $ln x - ln (\frac{1}{x}) = 2$ $ln x - ln (1/x) = 2$ ?

2 Answers

Explanation:

Explanation:

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How do you solve ln x - ln (1/x) = 2lnx−ln(1x)=2ln x - ln (1/x) = 2?

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How do you solve $ln x - ln (\frac{1}{x}) = 2$ $ln x - ln (1/x) = 2$ ?