Question 628331
There may be other ways to solve this but I think the easiest solution is based two keys:<ul><li>Recognizing that 4 is a power of 2 so it can be expressed as {{{2^2}}}. By doing this we can transform the equation so that the exponents have only one base: 2.</li><li>A good understanding of the rules of exponents. (You'll see what I mean by this shortly.)</li></ul>With these keys we can turn this equation into what is essentially a quadratic equation which can be solved using quadratic equation methods.<br>
First let's rewrite the 4:
{{{(2^2)^x+2^(x+1)-35 = 0}}}
which simplifies to:
{{{2^(2x)+2^(x+1)-35 = 0}}}
As I said, we are trying to get this to look like a quadratic equation. More precisely we are trying to put this equation into what is called "quadratic form". Quadratic form is where the exponent of the first term is twice the exponent of the middle term. The exponent of our first term is 2x. This is almost twice the exponent of the middle term. In fact it would be exactly twice if that +1 was gone.<br>
This is where a good understanding of the rules of exponents comes into play. What rule of exponents would cause exponents to be added? Answer: {{{a^p*a^q = a^(p+q)}}}. Usually we use this rule from left to right. But it also works going backwards! So we can rewrite {{{2^(x+1)}}} as {{{(2^x)(2^1)}}} or  {{{2*2^x}}}. Take a moment to let this sink in. Replacing {{{2^(x+1)}}} with {{{2*2^x}}} we get:
{{{2^(2x) + 2*2^x -35 - 0}}}
This is in quadratic form since the exponent of the first term, 2x, is exactly twice the exponent of our new middle term, x.<br>
Solving a quadratic form equation is done in much the same way as "regular" quadratic equations. But they can be difficult until you have had some practice. It can be helpful to use a temporary variable. Make the temporary variable equal to the base and exponent of the middle term. In this case:
Let {{{q = 2^x}}}
This makes {{{q^2 = (2^x)^2 = 2^(2x)}}}
Substituting these into our equation we get:
{{{q^2-2q-35=0}}}
This is obviously a quadratic equation which we can solve in the normal ways. This factors easily:
{{{(q-7)(q+5) = 0}}}
Using the Zero Product Property:
{{{q-7 = 0}}} or {{{q+5 = 0}}}
Solving these we get:
q = 7 or q = -5<br>
Of course we are not interested in solutions for our made-up variable q, We want solutions for x. So we substitute back in for q. (Remember, it was a <i>temporary</i> variable.)
{{{2^x = 7}}} or {{{2^x = -5}}}
Now we solve these. Since a power of 2 can never be negative, there is no solution to the second equation. But we can solve the first one. We use logarithms to solve this. For the simplest possible solution, use a logarithm with a base that matches the base of the exponent. In this case use base 2 logarithms:
{{{log(2, (2^x)) =log(2, (7))}}}
Using a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, we can move the exponent out front:
{{{x*log(2, (2)) =log(2, (7))}}}
And be definition, {{{log(2, (2)) = 1}}} so this simplifies to:
{{{x = log(2, (7))}}}
which is the exact solution to your equation.<br>
Any other base of logarithm can be used, too. You just end up with a slightly more complex expression. If you are interested in a decimal approximation, you might want to choose a base your calculator "knows", base 10 (log) or base e (ln):
{{{2^x = 7}}}
{{{ln(2^x) = ln(7)}}}
Use the property:
{{{x*ln(2)=ln(7)}}}
Here the ln(2) does not "disappear" like the {{{log(2, (2))}}} did earlier. This is why we end up with a more complex expression. Dividing both sides by ln(2):
{{{x=ln(7)/ln(2)}}}
This is another exact expression for the solution. It's a little more complex but it is easy to put into your calculator if you want/need a decimal approximation. (Note: if you do want a decimal, be sure to find the logarithms first and then divide. Do not divide 7 by 2 and then find a logarithm!)<br>
P.S. After some practice with these quadratic form equations you will not need a temporary variable. You will "see" how to go directly from
{{{2^(2x) + 2*2^x -35 - 0}}}
to
{{{(2^x-7)(2^x+5) = 0}}}
to
{{{2^x-7 = 0}}} or {{{2^x+5=0}}}