Question 372457
{{{2^((2/3)x+1) - 3*2^((1/3)x) - 20 = 0}}}
This is a tough one. The keys to solving this are:<ul><li>Recognizing that the first exponent is <i>almost</i> exactly twice the second exponent. (The +1 keeps it from being exactly twice.)</li><li>If the first exponent was twice the second one then this equation would be in "quadratic form" for {{{2^((1/3)x)}}} and quadratic form equations can be solved.</li><li>The +1 in the first exponent can be "factored out". Just like {{{2*2^4 = 2^(4+1)}}} {{{2*2^((2/3)x) = 2^((2/3)x + 1)}}}</li></ul>
So by factoring out a 2 from the first term we end up with a solvable quadratic form equation:
{{{2*2^((2/3)x) - 3*2^((1/3)x) - 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:
{{{(2*2^((1/3)x) + 5)(2^((1/3)x) - 4) = 0}}}
From the Zero Product Property we know that this (or any) product can be zero <i>only</i> if one (or more of the factors is zero. So:
{{{2*2^((1/3)x) + 5 = 0}}} or {{{2^((1/3)x) - 4 = 0}}}
Subtracting 5 from both sides of the first equation we get:
{{{2*2^((1/3)x) = -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^((1/3)x) = 4)}}}
The quick way to solve this is to realize that 4 is also a power of 2:
{{{2^((1/3)x) = 2^2)}}}
In order for these powers of 2 to be equal , the exponents must be equal:
{{{(1/3)x = 2)}}}
Multiply both sides by 3 we get:
x = 6
which is our solution.<br>
<b>"Using a temporary variable"</b>
We have an equation:
{{{2*2^((2/3)x) - 3*2^((1/3)x) - 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^((1/3)x)}}}
then {{{q^2 = (2^((1/3)x))^2 = 2^((2/3)x)}}}
Substituting these into the equation we get:
{{{2q^2 -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 <i>only</i> 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^((1/3)x) = (-5)/2}}} or {{{2^((1/3)x) = 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^((1/3)x) = 4)}}}
The quick way to solve this is to realize that 4 is also a power of 2:
{{{2^((1/3)x) = 2^2)}}}
In order for these powers of 2 to be equal, the exponents must be equal:
{{{(1/3)x = 2)}}}
Multiply both sides by 3 we get:
x = 6
which is our solution.<br>