Question

can someone help me solve this equation:
2^2x-2^x=8

Answer 1

actually my answer is wrong. solve
\[u^2-u-8=0\]

Answer 2

Let X = 2^x:
x^2 - x - 8 = 0
X = (1 +/- V33) /2
2^x = (1 +/- V33) /2
Take log2 both sides:
=> x = log2 (1 + V33) /2

Answer 3

= 1.75

Answer 4

\[2^{3.5}-2^{1.75}=7.9502\]
Close but is it close enough??

Answer 5

I think so, but I'm retired guvemint

Answer 6

Hi robtobey

Answer 7

No real root. Refer to the attachment.

Answer 8

Hi radar.

Answer 9

That is what Wolframalpha say also, a complex solution. Have a nice day robtobey