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.

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) (Show Source): You can put this solution on YOUR website!

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 and quadratic form equations can be solved.
The +1 in the first exponent can be "factored out". Just like

So by factoring out a 2 from the first term we end up with a solvable quadratic form equation:

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:

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:
or
Subtracting 5 from both sides of the first equation we get:

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:

The quick way to solve this is to realize that 4 is also a power of 2:

In order for these powers of 2 to be equal , the exponents must be equal:

Multiply both sides by 3 we get:
x = 6
which is our solution.

"Using a temporary variable"
We have an equation:

It often takes practice to see how to make the big step I used above. Until then you can use a temporary variable.
Let
then
Substituting these into the equation we get:

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:
or
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.

The quick way to solve this is to realize that 4 is also a power of 2:

In order for these powers of 2 to be equal, the exponents must be equal:

Multiply both sides by 3 we get:
x = 6
which is our solution.

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