4^(x-1) = 2^x +8
Any guiding tips would be greatly appreciated!

Question

4^(x-1) = 2^x +8
Any guiding tips would be greatly appreciated!

anonymous · Answer

This? $$4^{x-1} = 2^{x}+8$$

anonymous · Answer

You got it.

anonymous · Answer

you sure it is not 
$$4^{x+1}=2^{x+8}$$??

anonymous · Answer

4x-1+8

anonymous · Answer

or rather 
$$4^{x-1}=2^{x+8}$$

anonymous · Answer

Alex had it right. It's $$4^{x-1} = 2^x + 8$$

anonymous · Answer

Hmm, honestly I'm not sure how you would solve that...

zarkon · Answer

$$u=2^x$$

anonymous · Answer

2^(2x-2) - 2^x - 8 = 0
Or just 2^(2x-2) = 2^x + 2^3

anonymous · Answer

Not sure from there

turingtest · Answer

$$\frac14 (2^{2x})-2^x-8=0$$then use the quadratic formula perhaps...?

zarkon · Answer

yes...$$u=2^x$$  ;)

anonymous · Answer

i dont think so, it's exponential not quadratic.

anonymous · Answer

Zarkon what do you mean?

zarkon · Answer

it is quadratic in $2^x$

anonymous · Answer

How? What if x isn't 2?

turingtest · Answer

$$(2^x)^2-4(2^x)-32=0$$$$2^x=\frac{4\pm\sqrt{4^2+4(32)}}{2}=2\pm6$$

anonymous · Answer

How'd you do that?

turingtest · Answer

I listened to Zarkon...

anonymous · Answer

Well, i didnt know what he was saying in the first place

turingtest · Answer

I also used your formula
2^(2x-2) - 2^x - 8 = 0
and modified it a little...

anonymous · Answer

Will you show me?

turingtest · Answer

$$2^{2x-2} - 2^x - 8 =2^{-2}\cdot2^{2x}-2^x-8$$$$=\frac14(2^x)^2-2^x-8= 0$$multiply by 4 and you get what I wrote.

anonymous · Answer

Thanks Turing and Zarkon, I got it!
It works out once you substitute 2^x and use the guadratic formula.
(Don't forget to plug back into 2^x)

turingtest · Answer

All thanks to Zarkon, I had no idea how to solve it ;)

anonymous · Answer

Sweet turing, thanks, I still do not understand what zarkon was saying though
u = 2^x
is u just a random variable?

zarkon · Answer

random variable...only if x was a random variable ;)

turingtest · Answer

if you sub 2^x=u you get$$\frac14u^2-u-8=0$$so that sub makes the method clear

anonymous · Answer

Why couldn't he just say... I thought it was for some formula I'd never heard of, always used a,b,c for quadratic.

turingtest · Answer

Zarkon is not very chatty, you have to make the most of what he says :D

anonymous · Answer

I see. Thanks guys

jamesj · Answer

Yes, Zarkon is famously laconic.

anonymous · Answer

Perhaps, but helpful for sure! :D