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

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To solve the logarithmic equation log_2(x+2) + log_(x-2)4 = 3, the user attempts to manipulate the equation using properties of logarithms. They express the equation in terms of common logarithms and simplify it, leading to a complex expression involving log(x+2) and log(x-2). The user questions whether any value of x satisfies the equation, suggesting that their analysis indicates no solutions exist. Graphical analysis supports this conclusion, confirming that there are no valid x values that fulfill the equation. The discussion emphasizes the challenges of solving logarithmic equations with different bases and variables.
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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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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