Complicated logarithm problemor at least i think it is

[QuantumKing]

I was doing some assignment i have to give in, for math, and came upon this exponential equation: (2^x+1) + (2^x+2) = (2^1-x) + (2^3-x)

I thought, pfft, that's easy...so i did it, wrong answer, tried something else, wrong answer..tried another tactic, and i think you can guess what happened..Each answer i get is wrong and, to my judment, I am using a method of development that should work. but it doesnt. looked easy at first glance but isnt..lol, can anyone start me off on the right track?

 

[Mace]

hi

Hi my name is Mace, and i was woundering how do u post something like wat u did?

 

[QuantumKing]

haha, what do you mean?

 

[Mace]

Hi

put up a post?
can u tell me?
i am kinda new.

 

[Checkfate]

On the main forum page, press "New Topic" in the upper left. TO get to the main forum page from this page, scroll up and press "Introductory Physics". Welcome to PF

ps - I will work on the main question and get back to you in a minute.

 

[Checkfate]

Can you rewrite the equation as
2^{x+1}+3=10-2x ?

Does that help?

 

[QuantumKing]

ummm, how did you get that??

 

[Checkfate]

I should also explain the trick... it's kind of subtle.

If you have a common base and a common exponent and you want to add them, you MUST have the same number of terms as the base! If you wanted to add 3^{x} + 3^{x} + 3^{x} it would be 3^{x+1} but of course 3^x + 3^x can't be added. And if you wanted to add 4^x you would need four of them, etc. Kind of a useless trick really, but it helps here!

Perhaps someone else can explain why it works, I think it's probably simple but I just remember the trick, not the reason. lol

 

[QuantumKing]

but i don't see where your trick applies here, lol..i have two terms with the base=2..but different exponents.

 

[Checkfate]

lol! You have to use brackets! I thought it was 2^x + 1 not 2^(x+1) Sorry. lol

 

[0rthodontist]

If your equation is
2^(x+1) + 2^(x+2) = 2^(1-x) + 2^(3-x)
then you have
2^(x+1) + 2*2^(x+1) = 2^(1-x) + 4*2^(1-x)
Do you see what to do next?

 

[QuantumKing]

kewl kewl, got it thanks