SLIDE 3 Focusing on properties of the exponent
Whenever we work with exponential functions, we will eventually work with the inverse function. In preparation for that, here is an exercise which focuses on properties of the exponent. Suppose that 8a = 3 and 8b = 5. Find the exponent on 8 that gives
1 2 2 15 3 25 4 10
(Answers in (b), (c), and (d) will involve the unknowns a and/or b.)
- Solutions. Since 8a = 3 and 8b = 5 then
1 2 = 81/3. So our answer is 1/3 . 2 15 = (3)(5) = (8a)(8b) = 8a+b. So our answer is a + b . 3 25 = (5)2 = (8b)2 = 82b. So our answer is 2b . 4 10 = (2)(5) = (8
1 3 )(8b) = 8 1 3 +b. So our answer is
1 3 + b .
Smith (SHSU) Elementary Functions 2013 9 / 23
Focusing on properties of the exponent
Here are some sample problems from
- Dr. Paul’s online math notes on logarithms at Lamar University.
Example 1. Solve the following exponential equations for x.
1 53x = 57x−2 2 4t2 = 46−t
Solutions.
1 To solve 53x = 57x−2, we note that the bases are the same and so
(since f(x) = 5x is a one-to-one function) then we must have 3x = 7x − 2. This is a simple linear equation in x and a quick step or two leads to 4x = 2 so x = 1
2. 2 To solve 4t2 = 46−t, we again note that the bases are the same so
t2 = 6 − t. This is a quadratic equation in t. If we get zero on one side and write t2 + t − 6 = 0 we can factor this quadratic equation into (t + 3)(t − 2) = 0 and so t = −3 or t = 2.
Smith (SHSU) Elementary Functions 2013 10 / 23
Focusing on properties of the exponent
Continuing with Example 1.... Solve the following exponential equations for x.
3 3z = 9z+5
Solutions.
3 To solve 3z = 9z+5 in the same manner as before, we need to get the
bases to be equal. Let’s write 9 = 32 and make this problem one involving only base 3. So 3z = (32)z+5 and by properties of exponents 3z = 32(z+5). Therefore z = 2(z + 5). Now we have a simple linear equation. A step
Smith (SHSU) Elementary Functions 2013 11 / 23
Focusing on properties of the exponent
Continuing with Example 1.... Solve the following exponential equations for x.
3 45−9x = 1 8x−2
Solutions.
4 To solve 45−9x = 1 8x−2 we seek a common base. Let’s use base 2 and
write 4 = 22 and 8 = 23 so that the equation becomes (22)5−9x = 1 (23)x−2 . We then use properties of exponents to write 22(5−9x) = 1 23(x−2) . The expression on the righthand side (since 23(x−2) is in the denominator) can be rewritten as 22(5−9x) = 2−3(x−2). Now that our bases are the same, we solve the (easy!) linear equation 2(5 − 9x) = −3(x − 2) to find 4 = 15x so x = 4
15.
Smith (SHSU) Elementary Functions 2013 12 / 23