Expert: Sherman D. - 3/26/2007

Question
hi this is a problem i did. is my reasoning correct?

solve for x: 2^2x-3(2^x)-4=0
2^x*(2^x)-3(2^x)-4=0
2^x(2^x-3)=4
2^x(2^x-3)=2^2*2^0
2^x=2^2; x=2
ALSO (2^x-3)=2^0
    (2^x-3)=1
     2^x=4  ;x=2  

Answer
if by this you mean

2^(2x) - 3(2^x) - 4 = 0

instead of (2^x * 2^x) - 3(2^x) - 4 = 0, just write it like this (2^x)^2 - 3(2^x) - 4 = 0

same as saying

x^2 - 3x - 4 = 0
i just replaced "2^x" with "x", to make it more understandable.

you can use the quadratic formula for this

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-3) ± sqrt((-3)^2 - 4(1)(-4)))/(2(1))
x = (3 ± sqrt(9 + 16))/2
x = (3 ± sqrt(25))/2
x = (3 ± 5)/2
x = (-2/2) or (8/2)
x = -1 or 4

now we just put back the "2^x"

2^x = -1 or 2^x = 4

this can be rewritten as

x = log(2)(-1) or x = log(2)4

which can be written as

x = (log(-1))/(log(2)) or x = (log(4)/log(2))
x = Unknown or x = 2

since you can't actually have a negative log

ANS : x = 2

The way i have done it is much easier to do and understand.