2^(2x) - 2 ^ x - 6 = 0

[Screech]

I can get the answer to this from the back of the book ( log (base 2) 3 ) but I would prefer to know how they actually got this answer.

2^(2x) - 2 ^ x = 6

I've distributed a log throughout (not sure if you can do that but i did it anyway...) so that
2x log 2 - x log 2 = log 6
log ((2^2x)/(2^x) ) = log 6
log 2^x = log 6
x log 2 = log 6
x = log 6 / log 2
x = log (base 2) 6

However, it is supposed to be log base 2 of 3...

Also, I've multiplied all the powers to the (1/x) so that
2^2 - 2 = 6^(1/x)
2 = 6^(1/x)
log 2 = (1/x) log 6
x = log 6/log 2
x = log(base 2) 6

so, wtf am I consistently doing wrong?

 

[mzkhadir]

which is the problem ? Topic title has different one than the body

 

[fleshconsumed]

2^(2x) = 2^x*2^x

set y equal 2^x and solve as a quadratic eauation, then subsitite back and get your x

 

[Screech]

They're the same, i simply added the six on both sides for simplification.

 

[sciwizam]

Originally posted by: mzkhadir
which is the problem ? Topic title has different one than the body

Say whaaaa?

Look carefully, they are the same. 😛

 

[Goosemaster]

 

[mzkhadir]

Originally posted by: sciwizam

Originally posted by: mzkhadir
which is the problem ? Topic title has different one than the body

Say whaaaa?

Look carefully, they are the same. 😛

I know. I havent taken math in a while. Like the first replier said, he doesn't need to use log to solve the problem.

 

[Screech]

ahh, i get it now, thanks 🙂

 

[Goosemaster]

Originally posted by: fleshconsumed
2^(2x) = 2^x*2^x

set y equal 2^x and solve as a quadratic eauation, then subsitite back and get your x

:thumbsup:

 

[Random Variable]

2^x*2^x - 2 ^ x = 6

let z= 2^x
z^2 - z - 6 =0
(z-3)(z+2)=0

z= 3 or z = -2 (which you can ignore)

log(2)z=x

so x = log(2)3

 

[Random Variable]

I don't think you can take the log of the stuff of the left side of the equation individually.

 

[PurdueRy]

Originally posted by: Random Variable
I don't think you can take the log of the stuff of the left side of the equation individually.

This is correct.