SOLUTION: {{{16^x=4^(x^(2)-15)}}}

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Question 40284: 16%5Ex=4%5E%28x%5E%282%29-15%29
Answer by mbarugel(146) About Me  (Show Source):
You can put this solution on YOUR website!
Hello!
You must rewrite your equation so that the base number is equal on both sides of the equation. In your equation, on the left side, the base is 16, while on the thr right side, it's 4. However, notice that you could rewrite 16 as 4^2. So you would get:
+%284%5E2%29%5Ex+=+4%5E%28x%5E2-15%29
However, using the rule that says that +%28A%5EB%29%5EC+=+A%5E%28B%2AC%29, we get:
+4%5E%282x%29+=+4%5E%28x%5E2-15%29
Now, since the base is equal in both case, we simply remove it to get the equation:
2x+=+x%5E2+-+15
+0+=+x%5E2+-2x+-15
And now we simply solve a quadratic equation:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-15+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A-15=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+64+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+64+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-2%29-sqrt%28+64+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-2x%2B-15 can be factored:
1x%5E2%2B-2x%2B-15+=+%28x-5%29%2A%28x--3%29
Again, the answer is: 5, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-15+%29


So we get that the two solutions are X = 5 and X = -3

I hope this helps!
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