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Since the exponent of is exactly twice the exponent of , this equation is in what is called quadratic form. And these quadratic form equations can be solved in much the same way as regular quadratic equations.
To see the "quadratic-ness" of this equation it can help to use a temporary variable. Set it to the base and lower exponent. So:
Let .
Then .
Substituting these into our equation we get:
This is obviously a quadratic equation. It's already got a zero on one side. And it factors pretty easily:
(q-10)(q-3) = 0
From the Zero Product Property:
q-10 = 0 or q-3 = 0
Solving these:
q = 10 or q = 3
Of course we're not interested in solutions for q. We're interested in solutions for x. So we substitute back: or
We have a little more work to do to solve for x. The quick way is to cube both sides: or
which simplifies to:
x = 1000 or x = 27
P.S. Once you have done a few of these quadratic form equations you will no longer need a temporary variable. You will learn to see how to go directly from:
to
to or
etc.
P.S. Please put fractional exponents in parentheses.
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