SOLUTION: 2^(2/3)x+1 - 3*2^(1/3)x - 20 = 0. This would be read 2 to the (2/3)x+1[(2/3)x+1 is exponent], minus 3 times 2 to the (1/3)x [(1/3)x is exponent], minus 20 equals 0.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 2^(2/3)x+1 - 3*2^(1/3)x - 20 = 0. This would be read 2 to the (2/3)x+1[(2/3)x+1 is exponent], minus 3 times 2 to the (1/3)x [(1/3)x is exponent], minus 20 equals 0.      Log On


   

Question 372457: 2^(2/3)x+1 - 3*2^(1/3)x - 20 = 0.
This would be read 2 to the (2/3)x+1[(2/3)x+1 is exponent], minus 3 times 2 to the (1/3)x [(1/3)x is exponent], minus 20 equals 0.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E%28%282%2F3%29x%2B1%29+-+3%2A2%5E%28%281%2F3%29x%29+-+20+=+0
This is a tough one. The keys to solving this are:
  • Recognizing that the first exponent is almost exactly twice the second exponent. (The +1 keeps it from being exactly twice.)
  • If the first exponent was twice the second one then this equation would be in "quadratic form" for 2%5E%28%281%2F3%29x%29 and quadratic form equations can be solved.
  • The +1 in the first exponent can be "factored out". Just like 2%2A2%5E4+=+2%5E%284%2B1%29 2%2A2%5E%28%282%2F3%29x%29+=+2%5E%28%282%2F3%29x+%2B+1%29

So by factoring out a 2 from the first term we end up with a solvable quadratic form equation:
2%2A2%5E%28%282%2F3%29x%29+-+3%2A2%5E%28%281%2F3%29x%29+-+20+=+0
The next step is kind of a big one. If you have trouble following it, see "Using a temporary variable" below (where I use a lot of little steps instead of this one big one.) Next we factor the quadratic form equation:
%282%2A2%5E%28%281%2F3%29x%29+%2B+5%29%282%5E%28%281%2F3%29x%29+-+4%29+=+0
From the Zero Product Property we know that this (or any) product can be zero only if one (or more of the factors is zero. So:
2%2A2%5E%28%281%2F3%29x%29+%2B+5+=+0 or 2%5E%28%281%2F3%29x%29+-+4+=+0
Subtracting 5 from both sides of the first equation we get:
2%2A2%5E%28%281%2F3%29x%29+=+-5
But a power of 2 cannot be negative (and neither can 2 times a power of 2. So this equation has not solutions. But we still have the other equation. Adding 4 to each side of that equation we get:
2%5E%28%281%2F3%29x%29+=+4%29
The quick way to solve this is to realize that 4 is also a power of 2:
2%5E%28%281%2F3%29x%29+=+2%5E2%29
In order for these powers of 2 to be equal , the exponents must be equal:
%281%2F3%29x+=+2%29
Multiply both sides by 3 we get:
x = 6
which is our solution.

"Using a temporary variable"
We have an equation:
2%2A2%5E%28%282%2F3%29x%29+-+3%2A2%5E%28%281%2F3%29x%29+-+20+=+0
It often takes practice to see how to make the big step I used above. Until then you can use a temporary variable.
Let q+=+2%5E%28%281%2F3%29x%29
then q%5E2+=+%282%5E%28%281%2F3%29x%29%29%5E2+=+2%5E%28%282%2F3%29x%29
Substituting these into the equation we get:
2q%5E2+-3q+-+20+-+0
This is clearly a quadratic equation and it is not very hard to factor:
(2q + 5)(q - 4) = 0
From the Zero Product Property we know that this (or any) product can be zero only if one (or more of the factors is zero. So:
2q + 5 = 0 or q - 4 = 0
Solving these we get:
q = -5/2 or q = 4
We have found q. But we want to find x. So at this point we substitute back in for q:
2%5E%28%281%2F3%29x%29+=+%28-5%29%2F2 or 2%5E%28%281%2F3%29x%29+=+4
A power of 2 cannot be negative. So there are no solutions for the first equation. But we can find a solution to the second equation.
2%5E%28%281%2F3%29x%29+=+4%29
The quick way to solve this is to realize that 4 is also a power of 2:
2%5E%28%281%2F3%29x%29+=+2%5E2%29
In order for these powers of 2 to be equal, the exponents must be equal:
%281%2F3%29x+=+2%29
Multiply both sides by 3 we get:
x = 6
which is our solution.