[ASK] Logarithmic Equation

  • MHB
  • Thread starter Monoxdifly
  • Start date
  • #1
Monoxdifly
MHB
284
0
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?
 

Answers and Replies

  • #2
jonah1
108
0
Beer hangover induced idea follows.
20200915_141613.jpg
 
  • #3
Monoxdifly
MHB
284
0
So, no x fulfills the equation, right?
 
  • #4
skeeter
1,104
1
So, no x fulfills the equation, right?

That's what the graph says ...
 
  • #5
Monoxdifly
MHB
284
0
Okay, thanks guys.
 

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