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-- 3^x - 2*3^(1-x) = 1
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3^x - 2*3^(1-x) = 1

How does solve a beast such as this? I know it should be easy, but I just can't do it.

Any help appreciated.

FOIL


Idiot...

derive bitch, derive!

quote:

Originally posted by Boomer187 derive bitch, derive!

No... I don't think it has to be derived. Rather logarithms... but I can't figure it out atm.

quote:

Originally posted by Marc Summers FOIL Idiot...

we call that the "smiley face"

Didn't you take notes?

quote:

Originally posted by sensorium Didn't you take notes?

Help or stfu.

On a more serious note, yes, I did take notes, but we didn't go through this complicated versions of this. Or I just fail to see the connection.

What are your notes on? What subject?

quote:

Originally posted by sensorium What are your notes on? What subject?

I think it translates to "Exponential rules" or something like that. Exponents? Eh, I don't know.

quote:

Originally posted by sensorium What are your notes on? What subject?

Tic-Tac-Toe

quote:

Originally posted by Boomer187 derive bitch, derive!

MADD = mathematicians against drunk deriving.

dont drink and derive!

quote:

Originally posted by DJ Cinos I think it translates to "Exponential rules" or something like that. Exponents? Eh, I don't know.

You might be the best note taker ever.

Re: 3^x - 2*3^(1-x) = 1

quote:

Originally posted by DJ Cinos How does solve a beast such as this? I know it should be easy, but I just can't do it. Any help appreciated.

make the whole thing equal zero, so you just add a -1 to the equation. now put it in your calculator in the Y= part, and it'll form a graph. wherever that graph crosses that x-axis, thats your answer

quote:

Originally posted by sensorium You might be the best note taker ever.

No, but I don't take classes in English, smartass. They're called "Exponentiallagar" here, which directly translates to "Exponential Laws"

Re: Re: 3^x - 2*3^(1-x) = 1

quote:

Originally posted by PaperBag831 make the whole thing equal zero, so you just add a -1 to the equation. now put it in your calculator in the Y= part, and it'll form a graph. wherever that graph crosses that x-axis, thats your answer

The thing is, we can't use graph calculators. It'd be easy with one, yes... but we have to do it manually.

quote:

Originally posted by PaperBag831 MADD = mathematicians against drunk deriving. dont drink and derive!

omg... that joke was lame as cheese

(3^x) - (2) * (3^(1-x)) = 1

3^x * 3^(1-x) = 3

"3" is a like base.
figure the rest.



[edit]
whoops. that's going to give you 0.
open a textbook. it's a lot better for you.

Are you sure it works like that?

(hint: no)

EDIT: Oh, you noticed it yourself. Ah well, there aren't any examples of this in my books either. I'd better just concentrate on the other stuff.

quote:

Originally posted by Marc Summers omg... that joke was lame as cheese

its on a poster in the front of my math room

turn 3^x into log

I'm sure this will help you out somehow.

quote:

Originally posted by Yan I'm sure this will help you out somehow.

It's not about derivatives.

However, I found an entry on Exponents too, which COULD be of use.

Divide both sides by 3^(1-x),
this gives you 3^(2x-1) - 2 = 3^(x-1).

Rearrange to get (1/3)*3^2x - (1/3)*3^x - 2 = 0

Let y = 3^x, then solve as a regular quadratic equation.

This gives you y = -2, 3

3 = 3^x , so x = 1.

3^x - 6/3^x - 1 = 0
3^x - 2/1^x - 1 = 0 (1^x)
3^x^2 - 2 - 1^x = 0
3^x^2 - 1^x - 2 = 0
(3^x + 2)(1^x - 1) = 0

3^x = -2 1^x = 1

I forgot.

quote:

Originally posted by piggy Divide both sides by 3^(1-x), this gives you 3^(2x-1) - 2 = 3^(x-1). Rearrange to get (1/3)*3^2x - (1/3)*3^x - 2 = 0 Let y = 3^x, then solve as a regular quadratic equation. This gives you y = -2, 3 3 = 3^x , so x = 1.

Right. Thanks for the explanation. I still have trouble getting it, but that's just my own stupidity.

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