Question 481847
{{{2^x+2^(4-x)=17}}}
One way to solve this would be to start be rewriting the equation in terms of {{{2^x}}}. We can use the division rule for exponents, in reverse, on the second term to get:
{{{2^x+2^4/2^x=17}}}
The numerator simplifies to 16:
{{{2^x+16/2^x=17}}}
Now we can simplify the equation my eliminating the fraction. This can be done by multiplying both sides of the equation by {{{2^x}}}:
{{{2^x*(2^x+16/2^x)=2^x*(17)}}}
which simplifies to:
{{{2^(2x) + 16=17*2^x}}}
Next we get a zero on the right side by subtracting that term from each side:
{{{2^(2x) - 17*2^x + 16=0}}}
Since the exponent of {{{2^(2x)}}} is twice the exponent of the exponent in the middle term, {{{2^x}}}, this equation is in what is called "quadratic form". These kinds of equations can be solved just like "regular" quadratic equations. To make it easier to understand I am going to use a temporary variable:
Let {{{q = 2^x}}}
The {{{q^2 = (2^x)^2 = 2^(2x)}}}
Substituting these into the equation we get:
{{{q^2 - 17q +16 = 0}}}
This obviously a quadratic equation and factoring it is easy:
{{{(q-1)(q-16) = 0}}}
By the Zero Product Property:
q-1 = 0 or q-16 = 0
Solving these we get:
q = 1 or q = 16
We have solved for q. But we want to solve for x. Now we substitute for the temporary variable:
{{{2^x = 1}}} or {{{2^x = 16}}}
Only one power of 2 is equal to 1 and that is zero. So x = 0 is the only solution for the first equation. And there is only one power of 2 that is 16 and that is 4. So the only solution to the second equation is x = 4. Putting these together the complete solution to your equation is:
x = 0 or x = 4<br>
Once you have done a few "quadratic form" problems like this you will no longer need the temporary variable. You will see how to go from:
{{{2^(2x) - 17*2^x + 16=0}}}
to
{{{(2^x -1)(2^x - 16) = 0}}}
to
{{{2^x - 1 = 0}}} or {{{2^x - 16 - 0}}}
etc.