Question

HELP ME PLEASE (I WOULD TYPE THE QUESTION SHORTLY)

Answer 1

\[\huge IF a ^{2} -12ab +4b ^{2}=0\]
Prove that
\[\huge \log(a+2b)=\frac{ 1 }{ 2 }(loga +logb) +2\log2\]

Answer 2

@AravindG @ganeshie8 @AccessDenied @blues @Luigi0210 @Hero @zepdrix
@iPwnBunnies @wolf1728 @TheRealMeeeee

Answer 3

I'm not good at math man Sorry

Answer 4

NO PROBLEM

Answer 5

@dumbcow

Answer 6

@Hero We need your help

Answer 7

@zepdrix

Answer 8

thinking :3

Answer 9

I am too :)

Answer 10

I will be back wait!

Answer 11

\[\huge IF =if\]

Answer 12

"IF" may not be clear

Answer 13

If you apply some log rules, you can simplify the right side to this.
\[\Large\rm \log(a+2b)= \log 4\sqrt{ab}\]
Then exponentiating each side gives us:\[\Large\rm a+2b=4\sqrt{ab}\]And then square each side and you'll get the quadratic that we started with.

I guess the only problem is that we want to prove the log equation.
So we should start with the quadratic and do all of those steps in reverse.

Answer 14

I got it!

Answer 15

Understand what I'm saying?
Lemme finish up those steps in case there is any confusion.
Squaring each side,\[\Large\rm a^2+4ab+4b^2=16ab\]Subtracting 16ab from each side gives us:\[\Large\rm a^2-12ab+4b^2=0\]And then do all of those steps in reverse to prove the log equation.
Start by writing -12 as -16 and 4... etc.

Got it? :) cool.

Answer 16

yeah yeah
(a+2b)^2=16ab
after expand this stuff we get that quadratic
\[thank you \]

Answer 17

Expanding*