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[ASK] Logarithmic Equation

Monoxdifly

Well-known member
Aug 6, 2015
284
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?
 

jonah

New member
Feb 21, 2015
13
Beer hangover induced idea follows.
20200915_141613.jpg
 

Monoxdifly

Well-known member
Aug 6, 2015
284
So, no x fulfills the equation, right?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
702

Monoxdifly

Well-known member
Aug 6, 2015
284
Okay, thanks guys.