Math 115 / Section 1.3 (September 15, 2015)
page 4
6. Solve the following equations for x. Leave your answers in exact form.
a. aekx+b = Q, where a, k, b, and Q are positive consants
Solution: Dividing by a gives
ekx+b =
Q
.
a
Taking the logarithm of each side gives
ln ekx+b = ln
Q
.
a
kx + b = ln
Q
.
a
By the definition of ln, this is
Now finish solving for x:
1
x=
k
Q
ln
−b .
a
b. 5 · 3x = 2 · 10x
Solution: Taking the logarithm of each side gives
ln(5 · 3x ) = ln(2 · 10x ).
Using the fact that ln(ab) = ln a + ln b we have
ln 5 + ln 3x = ln 2 + ln 10x .
Using that ln ab = b ln a gives
ln 5 + x ln 3 = ln 2 + x ln 10.
Now finish solving for x:
x=
ln 5 − ln 2
.
ln 10 − ln 3
Math 115 / Section 1.3 (September 15, 2015)
c.
page 5
ln(8x) − 2 ln(2x)
=1
ln x
Solution: There are many ways to do this. One way:
ln(8x) − ln(4x2 )
=1
ln x
ln(8x) − ln(4x2 ) = ln x
8x
= ln x
ln
4x2
2
ln
= ln x
x
2
=x
x
2 = x2
√
This looks like it gives the solutions x = ±√ 2. But in the original equation, we have
terms like ln x. This is undefined for√x = − 2 since the domain of ln x is x > 0. So we
only keep the positive solution x = 2.
7. A scientist is growing bacteria for use in an experiment. Suppose the number of bacteria cells
in the culture is modeled by B(t) = 12(2t/31.3 ), where B(t) is the number of cells (in millions)
t hours after he begins.
a. How long does it take for the number of cells to double?
Solution: It takes 31.3 days for the number of cells to double. We can see this from the
formula; every time t increases by 31.3, the exponent on the 2 increases by 1.
You could also solve the equation B(t) = 24 for t.
b. By what percent does the number of cells grow each hour?
Solution: B(t) = 12(2t/31.3 ) = 12(21/31.3 )t ≈ 12(1.02239)t
So the number of cells grows by 2.239% each hour.
c. What is the continuous hourly percent growth rate of the number of cells?
Solution:
We need to convert B(t) = 12(2t/31.3 ) to the form B(t) = aekt . Since
B(0) = 12, we know a = 12.
12(2t/31.3 ) = 12ekt
(21/31.3 )t = (ek )t
Then we get
k = ln 21/31.3 ≈ 0.022145
so the continuous hourly percent growth rate is about 2.2145%.
Math 115 / Section 1.3 (September 15, 2015)
page 6
d. When will the scientist have 50 million bacteria cells?
Solution: We want to solve B(t) = 50.
50 = 12(2t/31.3 )
50
= 2t/31.3
12
50
ln
= ln 2t/31.3
12
50
t
ln
=
ln 2
12
31.3
31.3 ln 50
12
t=
≈ 64.44
ln 2
The scientist will have 50 million cells after about 64.44 hours.