How do you solve $ln x = 2 (ln 1 - ln 11)$ $ln x = 2(ln 1 - ln 11)$ ?

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Answer 1 · 2016-01-17T04:35:57

$x = \frac{1}{121}$ $x=1/121$

Divide both sides by $2$ $2$ .

$(\frac{1}{2}) ln x = ln 1 - ln 11$ $(1/2)lnx=ln1-ln11$

Simplify the right hand side using the logarithm rule: $ln a - ln b = ln (\frac{a}{b})$ $lna-lnb=ln(a/b)$

$(\frac{1}{2}) ln x = ln (\frac{1}{11})$ $(1/2)lnx=ln(1/11)$

Simplify the left hand side using the logarithm rule: $a ln x = ln (x^{a})$ $alnx=ln(x^a)$

$ln (x^{\frac{1}{2}}) = ln (\frac{1}{11})$ $ln(x^(1/2))=ln(1/11)$

$ln \sqrt{x} = ln (\frac{1}{11})$ $lnsqrtx=ln(1/11)$

Thus, since if $ln a = ln b$ $lna=lnb$ , then $a = b$ $a=b$ ,

$\sqrt{x} = \frac{1}{11}$ $sqrtx=1/11$

$x = {(\frac{1}{11})}^{2}$ $x=(1/11)^2$

$x = \frac{1}{121}$ $x=1/121$

How do you solve lnx=2(ln1−ln11)ln x = 2(ln 1 - ln 11)?