pre-cal question, solve algebraically for x?

Question

2^(2x)-6*(times)2^x=16

Answer 1

2^(2x) - 6•2^x = 16. This is quadratic. Replace 2^x with something else, w.

w² - 6w - 16 = 0
(w - 8)(w + 2) = 0
w = 8 ........ w = -2
2^x = 8 ....... 2^x = -2
x = 3 .............no solution

So x=3

Answer 2

When you multiply bases, you add the exponents

Change the 16 to 2^4
your equation then is equal to

2^2x-6 times 2 ^x = 2^4

now all the bases are the same,

2x-6+x = 4

unless I missed something in translating what you wrote to what I interpret,
3x = 10
x = 10/3

and it checks

Answer 3

well x^2 times x^2 = x^4, so multiplying two of the same numbers with exponents you just add the exponents so...
[2^(2x-6)]*2^x = 2^(3x-6) = 16

now you have to take the base 2 log of each side, I couldnt figure out how to do this on my calculator but that okay because taking the base 2 log of something means that we want to know what exponent we would need on 2 to get our answer (so what power do we raise 2 to in order to get 16?....2^4=16.....so 4


(log2)2^(3x-6)=(log2)16 = 3x-6=4 x=10/3