# Confused over application of log power rule ?

## The Question says In x+In2x=12 I got the right answer by e^(2x^2)=e^12 which gives 218.39 But when I try it with log power rule, Inx+In2x=12 In2x^2=12 2In2x=12 In2x=6 e^(In2x)=e^6,which gives me the wrong answer of 201.7. Why is this ?

Aug 14, 2016

Correct is ${e}^{6} / \sqrt{2} \approx 285.27$

#### Explanation:

$\ln x + \ln 2 x = 12$

I got the right answer by ${\textcolor{red}{e}}^{2 {x}^{2}} = {e}^{12}$ which gives 218.39

No, red bit is wrong

$\ln x + \ln 2 x = 12$
$\implies \ln 2 {x}^{2} = 12$
$\implies {e}^{\ln 2 {x}^{2}} = {e}^{12}$
$\implies 2 {x}^{2} = {e}^{12}$ NOT ${e}^{2 {x}^{2}} = {e}^{12}$
$\implies {x}^{2} = {e}^{12} / 2$
$\implies x = {\left({e}^{12} / 2\right)}^{\frac{1}{2}} = {e}^{6} / \sqrt{2}$

But when I try it with log power rule,
$\ln x + \ln 2 x = 12$
$\ln 2 {x}^{2} = 12$
$2 \ln 2 x = 12$ Nah, sorry

Correct is
$\ln 2 {x}^{2} = 12$
$\ln {\left(\sqrt{2} x\right)}^{2} = 12$
$2 \ln \left(\sqrt{2} x\right) = 12$
$\ln \left(\sqrt{2} x\right) = 6$
${e}^{\ln \left(\sqrt{2} x\right)} = {e}^{6}$
$\sqrt{2} x = {e}^{6}$
$x = {e}^{6} / \sqrt{2}$