
Logarithmic Equation
$\displaystyle 2[\log_{2}(x+5)]^2 + \log_{2}(x5)\log_{2}(x+5) = [\log_{4}(x^210+25)]^2$
I was able to simplify until it became
$\displaystyle 2[\log_{2}(x+5)]^2 + \log_{2}(x5)\log_{2}(x+5) = [\frac{\log_{2}(x5)^2}{2}]^2$
Then I said: Let $\displaystyle a= \log_{2}(x+5)$ and let $\displaystyle b= \log_{2}(x5)$
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.
Thanks in advance.

Re: Logarithmic Equation
What is the original problem in its entirety?

Re: Logarithmic Equation
The original problem is the first equation.

Re: Logarithmic Equation
You may factor $\displaystyle 2a^2+abb^2$ as $\displaystyle (a+b)(2ab)$.

Re: Logarithmic Equation
Thank you, but I'm still unsure about where to go from there. How can we find a value for x?

Re: Logarithmic Equation
This gives you two equations using the zerofactor property:
i) $\displaystyle a+b=0$
$\displaystyle \log_2(x+5)+\log_2(x5)=0$
$\displaystyle \log_2(x^225)=0$
$\displaystyle x^225=1$
ii) $\displaystyle 2ab=0$
$\displaystyle 2\log_2(x+5)=\log_2(x5)$
$\displaystyle (x+5)^2=x5$
Now, solve the two equations for real roots, and keep in mind we require $\displaystyle 5<x$.

Re: Logarithmic Equation
Oh. I completely forgot about that property. Thank you. :)