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QUESTION: If 8(4^x-2^x)-2=2^(x+1) find x thanks
ANSWER: (1) Remember, in this solution, that ln(b)/ln(a) = log(b) base a.
(2) Also note 4^x = (2^x)(2^x) = 2^(2x).
Take 8(4^x-2^x)-2=2^(x+1) and divide both sides by 2.
This give 4(4^x-2^x) - 1 = 2^x.
Next, multiply the left side out, giving 4*4^x - 4*2^x - 1 = 2^x.
Subtract 2^x from both sides, giving 4*4^x - 5*2^x - 1 = 0.
Setting a = 2^x and using (2) above, we get 4a^2 - 5a - 1 = 0.
Using the quadratic equation, a = (5 +/- sqrt(25+16))/8.
Since 25+16=41, sqrt(41) > 5, we must have (5 + sqrt(25+16))/8.
This is since subtracting off the sqrt(41) would give a negative value for a,
which is not possible since a = 2^x and we are looking for a real solution.
Since a = 2^x, ln(a) = x*ln(2), so x = ln(a)/ln(2).
This means x = log(a), base 2, where a = (5 + sqrt(41))/8.
This means we can also say x = log(5 + sqrt(41)) - 3, since log(8) = 3.
---------- FOLLOW-UP ----------
QUESTION: I don't understand and the answer provided in the txtbook is 0 and -2
The equation is 8(4^x - 2^x) - 2 = 2^(x+1).
Putting in x=0 gives 8(1-1) - 2 = 1, which is -2 = 1, and that is wrong,
and 8(1/16 - 1/4) - 2 = 1/4, which is -24/17 - 2 = 1/4, and that is also wrong.
From these solutions, shouldn't the -2 be a +2?
That is, change the equation to 8(4^x - 2^x) + 2 = 2^(x+1).
This has solutions 0 and -2.
To see this, notice that all terms can be written as powers of 2.
That is, 4 = 2^2, so 4^x = 2^(2x).
This makes the equation 8(2^(2x) - 2^x) + 2 = 2^(x+1).
Dividing each term by 2 gives 4(2^(2x) - 2^x) + 1 = 2^x.
Multiplying out the term on the left gives 4*(2^[2x]) - 4*(2^x) + 1 = 2^x.
Subtract 2^x from both sides gives 4*(2^[2x]) - 5*(2^x) + 1 = 0.
If A = 2^x, this is 4A^2 - 5A + 1 = 0.
That can be factored to (4A - 1)(A - 1) = 0.
For this to be 0, we need 4A - 1 = 0 or A - 1 = 0.
This means A = 1/4 or A = 1.
Since A = 2^x, if A = 1/4, x = -2, and if A = 1, x = 0.
As I was taking math in college, occasionally we would run into an error in the book where a sign was changed or left out.
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