Using logs: Find the possible values of x for which :?

Question

2^(2x+1) = 3(2^x)-1

Answer 1

Previous answerer has an error. It should go like this:
2*(2^x)^2 = 3*(2^x) - 1.......(1)
Make temporary substitute
y = 2^x..........(2).
Then (1) becomes:
2*y^2 - 3*y + 1 = 0, which has two real roots:
y1 = (3 + 1)/4 = 1, and
y2 = (3 - 1)/4 = 1/2.
Coming back to (2) you get:
2^x1 = 1 => x1 = 0, and
2^x2 = 2^(-1) => x2 = -1.
Inverting (2) you can use x*ln(2) = ln(y), however it is obvious even without it.

Answer 2

Usually you can use logs to bring down expressions from the exponents, but in this case it's tricky because here you still have that "-1" term you can't really get rid of. So try this instead:

2^(2x+1) = 3(2^x) - 1
2^(2x) * 2 = 3(2^x) - 1
2 * (2^(x))^2= 3(2^x) - 1
2(2^(x))^2 - 3(2^x) + 1 = 0
Let p = 2^(x). Then this becomes
2p^2 - 3p + 1 = 0
(2p - 1)(p - 1) = 0
So p =1/2 or p = 1
this means x = -1 or x = 0

Answer 3

2^2x *2 =3*2^x-1
If you call 2^x=z you have
2 z^2-3z+1=0
z=((3+-sqrt(9-8))/4 so z=1 and z= 1/2
2^x =1 so x= 0
and 2^x=1/2 so x=-1

Answer 4

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Answer 5

2 * 2^x = 3 * 2^x -1
so 2^x = 1
so x = 0