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Arkansas Tech University MATH 2243: Business Calculus Dr. Marcel B. Finan
6 The Natural Logarithm
An equation of the form ax^ = b can be solved graphically. That is, using a calculator we graph the horizontal line y = b and the exponential function y = ax^ and then find the point of intersection. In this section we discuss an algebraic way to solve equations of the form ax^ = b where a and b are positive constants. For this, we introduce a function that is found in today’s calculators, namely, the function ln x.
y = ln x if and only if ey^ = x.
where e = 2. 71828 · · · We call ln x the natural logarithm of x.
Properties of Logarithms
(i) Since ex^ = ex^ we can write
ln ex^ = x
(ii) Since ln x = ln x we can write
eln^ x^ = x
(iii) ln 1 = 0 since e^0 = 1. (iv) ln e = 1 since e^1 = e. (v) Suppose that m = ln a and n = ln b. Then a = em^ and b = en. Thus, a · b = em^ · en^ = em+n. Rewriting this using logs instead of exponents, we see that ln (a · b) = m + n = ln a + ln b.
(vi) If, in (v), instead of multiplying we divide, that is ab = e m en^ =^ e
m−n (^) then
using logs again we find
ln
( (^) a
b
) = ln a − ln b.
(vii) It follows from (vi) that if a = b then ln a − ln b = ln 1 = 0 that is ln a = ln b. (viii) Now, if n = ln b then b = en. Taking both sides to the power k we find bk^ = (en)k^ = enk. Using logs instead of exponents we see that ln bk^ = nk = k ln b that is ln bk^ = k ln b.
Example 6. Solve the equation: 4(1.171)x^ = 7(1.088)x.
Solution. Rewriting the equation into the form
( (^1). 171
- 088
)x = 74 and then using properties
(vii) and (viii) to obtain
x ln
(
- 171
- 088
) = ln
Thus,
x =
ln (^74) ln
(
- 171
- 088
Example 6. Solve the equation ln (2x + 1) + 3 = 0.
Solution. Subtract 3 from both sides to obtain ln (2x + 1) = − 3. Switch to exponential form to get 2x + 1 = e−^3. Subtract 1 and then divide by 2 to obtain x = e−^3 − 1 2 ≈ −^0.^4995
Remark 6. Keep in mind the following:
- ln (a + b) 6 = ln a + ln b. For example, ln 2 6 = ln 1 + ln 1 = 0.
- ln (a − b) 6 = ln a − ln b. For example, ln (2 − 1) = ln 1 = 0 whereas ln 2 − ln 1 = ln 2 6 = 0.
- ln (ab) 6 = ln a · ln b. For example, ln 1 = ln (2 · 12 ) = 0 whereas ln 2 · ln 12 = − ln^2 2 6 = 0.
- ln
( a b
) 6 = ln ln^ a b. For example, letting a = b = 2 we find that ln ab = ln 1 = 0
whereas ln ln^ ab = ln 2ln 2 = 1.
( 1 a
) 6 = (^) ln^1 a. For example, ln (^11) 2
= ln 2 whereas (^) ln^1 2
= − (^) ln 2^1.
Solution. We are given the initial value 7.3 million and the continuous growth rate k = 0. 022. Therefore, P (t) = 7. 3 e^0.^022 t. Next,we want to find the time when P (t) = 10. That is , 7. 3 e^0.^022 t^ = 10. Divide both sides by 7.3 to obtain e^0.^022 t^ ≈ 1. 37. Solving this equation to obtain t = ln 1 0. 022.^37 ≈ 14. 3
Next, in order to convert from Q(t) = bekt^ to Q(t) = bat^ we let a = ek. For example, to convert the formula Q(t) = 7e^0.^3 t^ to the form Q(t) = bat^ we let b = 7 and a = e^0.^3 ≈ 1. 35. Thus, Q(t) = 7(1.35)t.
Example 6. Find the annual percent rate and the continuous decay rate of Q(t) = 200(0.886)t.
Solution. The annual percent rate of decrease is r = b − 1 = 0. 886 − 1 = − 0 .114 = − 11 .4%. To find the continuous decay rate we let ek^ = 0.886 and solve for k to obtain k = ln 0. 886 ≈ − 0 .121 = − 12 .1%
