How Do I Solve a Logarithmic Equation with Different Bases and Variables?

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The discussion centers on solving the logarithmic equation $$log_2(x+2)+log_{(x-2)}4=3$$. The user attempts to manipulate the equation using properties of logarithms, ultimately expressing it in terms of logarithmic identities. The conclusion drawn from the analysis indicates that there are no values of x that satisfy the equation, as confirmed by graphical representation.

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Monoxdifly
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A friend asked me how to solve this question:
$$log_2(x+2)+log_{(x-2)}4=3$$
I said I had no idea because one is x + 2 and the other one is x - 2. If both are x + 2 or x - 2, I can do it. He said that if that's the case, even at his level he could solve it. This is what I've done so far regarding the question.
$$log_2(x+2)+log_{(x-2)}4=3$$
$$\frac{log(x+2)}{log2}+\frac{log4}{log(x-2)}=3$$
$$\frac{log(x+2)log(x-2)+log4log2}{log2log(x-2)}=3$$
$$log(x+2)log(x-2)+2log^22=3log2log(x-2)$$
$$log(x+2)log(x-2)-3log2log(x-2)=-2log^22$$
$$log(x-2)(log(x+2)-3log2)=-2log^22$$
What should I do from here? Or did I make some mistakes?
 
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Beer hangover induced idea follows.
20200915_141613.jpg
 
So, no x fulfills the equation, right?
 
Monoxdifly said:
So, no x fulfills the equation, right?

That's what the graph says ...
 
Okay, thanks guys.
 

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