SOLUTION: Solve the following exponential functions a. 5 (with an exponent of) 5x+1 = 125 b. 3 (with an exponent of x squared) = 81 (with an exponent of x)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve the following exponential functions a. 5 (with an exponent of) 5x+1 = 125 b. 3 (with an exponent of x squared) = 81 (with an exponent of x)      Log On


   

Question 380913: Solve the following exponential functions
a. 5 (with an exponent of) 5x+1 = 125
b. 3 (with an exponent of x squared) = 81 (with an exponent of x)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
With many equations where the variable is in an exponent, one usually uses logarithms to solve the equation. An exception is when both sides of the equation can be expressed as powers of the same number. Both of your equations fit into this later category.

a. 5%5E%285x%2B1%29+=+125
Since 125+=+5%5E3 we can write this as:
5%5E%285x%2B1%29+=+5%5E3
Both sides are not powers of the same number, 5. In order for these powers of 5 to be equal, the exponents must be equal:
5x+1 = 3
This is a simple equation to solve:
5x = 2
x = 2/5

P.S. As you noted in your thank-you note, I overlooked the exponent on the 81 in the second problem. Here's the corrected solution:
b. 3%5E%28%28x%5E2%29%29+=+81%5Ex
Since 81+=+3%5E4 we can rewrite this as:
3%5E%28%28x%5E2%29%29+=+%283%5E4%29%5Ex
Using the power of a power rule for exponents (i.e. multiply) we get:
3%5E%28%28x%5E2%29%29+=+3%5E%284x%29
Again this equation is true only if the exponents are equal:
x%5E2+=+4x
This is a quadratic equation. To solve it we need one side equal to zero. Subtracting 4x from each side we get:
x%5E2+-+4x+=+0
Now we factor (or use the Quadratic Formula). This factors very simply:
x(x-4) = 0
From the zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
x = 0 or x-4 = 0
Solving the second equation we get:
x = 0 or x = 4