[ASK] Logarithmic Equation

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  • Thread starter Monoxdifly
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    Logarithmic
In summary, the conversation was about solving the equation log_2(x+2)+log_{(x-2)}4=3. The speaker admitted to not knowing how to solve it due to the terms x+2 and x-2 being different. The friend mentioned that if they were the same, he could solve it easily. The speaker then showed their progress in solving the equation and asked for further guidance. The conversation ended with the consensus that no x value satisfies the equation.
  • #1
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A friend asked me how to solve this question:
\(\displaystyle log_2(x+2)+log_{(x-2)}4=3\)
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.
\(\displaystyle log_2(x+2)+log_{(x-2)}4=3\)
\(\displaystyle \frac{log(x+2)}{log2}+\frac{log4}{log(x-2)}=3\)
\(\displaystyle \frac{log(x+2)log(x-2)+log4log2}{log2log(x-2)}=3\)
\(\displaystyle log(x+2)log(x-2)+2log^22=3log2log(x-2)\)
\(\displaystyle log(x+2)log(x-2)-3log2log(x-2)=-2log^22\)
\(\displaystyle log(x-2)(log(x+2)-3log2)=-2log^22\)
What should I do from here? Or did I make some mistakes?
 
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  • #2
Beer hangover induced idea follows.
20200915_141613.jpg
 
  • #3
So, no x fulfills the equation, right?
 
  • #4
Monoxdifly said:
So, no x fulfills the equation, right?

That's what the graph says ...
 
  • #5
Okay, thanks guys.
 

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