# Having Trouble Solving Logarithmic Equation

• October 17th 2011, 03:51 PM
Algebrah
Having Trouble Solving Logarithmic Equation
I'm having some trouble with this question. First I have to solve the following equation for $x$, then I have to substitute the value into $g(x) = 5(3x^2)^x$. Help would be greatly appreciated!

Solving For X

$\ln(5x^2) + 6 = 0$

$\ln(5x^2) = -6$

$e^{\ln(5x^2)} = e^{-6}$

$5x^2 = e^{-6}$

$x^2 = \frac{e^{-6}}{5}$

$\sqrt{x^2} = \frac{\sqrt{e^{-6}}}{\sqrt{5}}$

$x = \frac{1}{e^3\sqrt{5}}$

Substitution

$g(x) = 5(3x^2)^x$

$g(\frac{1}{e^3\sqrt{5}}) = 5(3x^2)^x$

$g(\frac{1}{e^3\sqrt{5}}) = 5(3(\frac{1}{e^3\sqrt{5}})^2)^{\frac{1}{e^3\sqrt{5 }}}$

$g(\frac{1}{e^3\sqrt{5}}) = 5((\frac{3}{e^3\sqrt{5}})^2)^{\frac{1}{e^3\sqrt{5} }}$

$g(\frac{1}{e^3\sqrt{5}}) = 5(\frac{9}{(e^3\sqrt{5})^2})^{\frac{1}{e^3\sqrt{5} }}$

I'm stuck at this point (though I'm not sure if it's correct up to this point either (Lipssealed)).

• October 17th 2011, 04:27 PM
skeeter
Re: Having Trouble Solving Logarithmic Equation
I agree with your solution for x.

my evaluation for $g(x) = 5(3x^2)^x$ ...

$g\left(\frac{1}{e^3 \sqrt{5}}\right) = 5\left(\frac{3}{5e^6}\right)^{\frac{1}{e^{3} \sqrt{5}}}$
• October 17th 2011, 04:40 PM
Algebrah
Re: Having Trouble Solving Logarithmic Equation