Question 726405
A quadratic equation in standard form is {{{y=ax^2+bx+c}}}. In this equation the first exponent is twice the exponent in the middle. An equation where the first exponent is twice the middle exponent but the exponents are not on x's is called an equation in "quadratic form". Quadratic form equations can be solved in much the same way as regular quadratic equations.<br>
{{{3^(2x)+3^(x+1)-4=0}}}
If the "x+1" exponent was just "x", then your equation would be in quadratic form (since 2x is twice x). So our first task is to change that exponent into just "x". To do this we will do something you may not have done before. The "x+1" is an addition. And there is a rule for exponents that tells us when to add exponents:
{{{a^m*a^n = a^(m+n)}}}
What we are going to do is to use this rule to "un-add" the exponents. IOW, we are going to use this rule "backwards". Looking at {{{3^(x+1)}}} and thinking: "What powers of three would we have multiplied to end up with {{{3^(x+1)}}}?" I hope that you can see that it would be {{{3^x*3^1}}}. Rewriting that middle term this way we get:
{{{3^(2x)+3^x*3^1-4=0}}}
Since {{{3^1 = 3}}} and since it is multiplication we can change the order, we can rewrite the middle term to:
{{{3^(2x)+3*3^x-4=0}}}
The equation is now in quadratic form.<br>
If you have not solved many quadratic form equations, then it can be difficult to see how to proceed. It can be helpful to use a temporary variable. Set it equal to the base and the lower exponent. So... let {{{q = 3^x}}}, then {{{q^2 = (3^x)^2 = 3^(2x)}}}. Substituting these into the equation we get:
{{{q^2+3q-4=0}}}
This equation is clearly quadratic. And we can see how to factor it:
{{{(q-1)(q+4) = 0}}}
From the Zero Product Property we get:
q-1 = 0 or q+4 = 0
Solving these we get:
q = 1 or q = -4<br>
Of course we are not interested in solutions for our made-up variable q. So we now substitute back in for the q's:
{{{3^x = 1}}} or {{{3^x = -4}}}
To solve for x we have a little more work to do. Since no power of 3 would ever be negative, the second equation will have no solutions. For the first equation we have to figure out what power of 3 results in a 1. Fortunately, if we know our exponents, we know that the exponent we are looking for is zero! So the only solution to the equation is x = 0. (If it had not been a well-known power of 3 then we would have to use logarithms to find x.)<br>
P.S. Once you have done a few of these quadratic form equations you will no longer need a temporary variable. You will see how to go directly from:
{{{3^(2x)+3*3^x-4=0}}}
to
{{{(3^x-1)(3^x+4) = 0}}}
to
{{{3^x-1 = 0}}} or {{{3^x+4 = 0}}}
etc.