Logarithmic Equation

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November 20th 2012, 07:54 PM
miguel11795

Logarithmic Equation

$2[\log_{2}(x+5)]^2 + \log_{2}(x-5)\log_{2}(x+5) = [\log_{4}(x^2-10+25)]^2$

I was able to simplify until it became

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

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

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

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.
November 20th 2012, 08:01 PM
MarkFL

Re: Logarithmic Equation

What is the original problem in its entirety?
November 20th 2012, 08:12 PM
miguel11795

Re: Logarithmic Equation

The original problem is the first equation.
November 20th 2012, 08:31 PM
MarkFL

Re: Logarithmic Equation

You may factor $2a^2+ab-b^2$ as $(a+b)(2a-b)$ .
November 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?
November 20th 2012, 08:57 PM
MarkFL

Re: Logarithmic Equation

This gives you two equations using the zero-factor property:

i) $a+b=0$

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

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

$x^2-25=1$

ii) $2a-b=0$

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

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

Now, solve the two equations for real roots, and keep in mind we require $5<x$ .
November 20th 2012, 09:05 PM
miguel11795

Re: Logarithmic Equation

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

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