# Logarithmic Equation

• Nov 20th 2012, 07:54 PM
miguel11795
Logarithmic Equation
$\displaystyle 2[\log_{2}(x+5)]^2 + \log_{2}(x-5)\log_{2}(x+5) = [\log_{4}(x^2-10+25)]^2$

I was able to simplify until it became

$\displaystyle 2[\log_{2}(x+5)]^2 + \log_{2}(x-5)\log_{2}(x+5) = [\frac{\log_{2}(x-5)^2}{2}]^2$

Then I said: Let $\displaystyle a= \log_{2}(x+5)$ and let $\displaystyle b= \log_{2}(x-5)$

so it becomes $\displaystyle 2a^2 + ab = (\frac{2b}{2})^2$

I don't know where to go from there. I'm trying to get a system of equation, but I can't seem to get another equation.

• Nov 20th 2012, 08:01 PM
MarkFL
Re: Logarithmic Equation
What is the original problem in its entirety?
• Nov 20th 2012, 08:12 PM
miguel11795
Re: Logarithmic Equation
The original problem is the first equation.
• Nov 20th 2012, 08:31 PM
MarkFL
Re: Logarithmic Equation
You may factor $\displaystyle 2a^2+ab-b^2$ as $\displaystyle (a+b)(2a-b)$.
• Nov 20th 2012, 08:48 PM
miguel11795
Re: Logarithmic Equation
Thank you, but I'm still unsure about where to go from there. How can we find a value for x?
• Nov 20th 2012, 08:57 PM
MarkFL
Re: Logarithmic Equation
This gives you two equations using the zero-factor property:

i) $\displaystyle a+b=0$

$\displaystyle \log_2(x+5)+\log_2(x-5)=0$

$\displaystyle \log_2(x^2-25)=0$

$\displaystyle x^2-25=1$

ii) $\displaystyle 2a-b=0$

$\displaystyle 2\log_2(x+5)=\log_2(x-5)$

$\displaystyle (x+5)^2=x-5$

Now, solve the two equations for real roots, and keep in mind we require $\displaystyle 5<x$.
• Nov 20th 2012, 09:05 PM
miguel11795
Re: Logarithmic Equation
Oh. I completely forgot about that property. Thank you. :)