BCUSP 123A Section 4.1 & 4.2: Logs and Using Logs to Solve Exponential Equations Solutions
1.3
1. Given 10
≈ 20, approximate log 200 without using a calculator.
1.3
By definition, 10 ≈ 20 means log 20 ≈ 1.3
Need a log 20 to solve this. So, log 200 = log (10·20) = log 10 + log 20 = 1 + 1.3 = 2.3
x
2. Solve the equation for x: Pa = Qb
x
Can solve this in many different ways. Either first isolate the terms with x in them, and then take logs of both
sides. Or you can take logs of both sides first.
Way #1 to solve it (note that the x only goes with a and b):
x
x
Pa = Qb
x
x
log (Pa ) = log (Qb )
take log of both sides
x
x
log P + log a = log Q + log b
use property: log (ab) = log a + log b
x
log P + x log a = log Q + x log b
use property: log a = x log a
x log a – x log b = log Q – log P
put terms with x on the same side, constants on other side
x(log a – log b) = log Q – log P
factor out x
x = (log Q – log P)/(log a – log b)
divide both sides by the value (log a – log b)
Way #2 to solve it:
x
x
Pa = Qb
x x
a /b = Q/P
x x
log (a /b ) = log (Q/P)
x
x
log a – log b = log (Q/P)
x log a – x log b = log (Q/P)
x(log a – log b) = log (Q/P)
x = log (Q/P)/log a – log b)
x
divide both sides by P and by b
take logs of both sides
use property: log (a/b) = log a – log b
x
use property: log a = x log a
factor out x
divide both sides by the value (log a – log b)
Notice that log (Q/P) = log Q – log P
3. Solve for x down to logs: 58 e
58 e 4x+1 = 30
ln (58 e 4x+1) = ln 30
ln 58 + ln (e 4x+1) = ln 30
ln 58 + (4x+1) ln e = ln 30
ln 58 + 4x+1 = ln 30
4x = ln 30 – ln 58 – 1
x = (ln 30 – ln 58 – 1)/4
4x+1
= 30
take natural logs of both sides
use property: ln (ab) = ln a + ln b
use property: ln ax = x ln a
since ln e = 1
subtract ln 58 and 1 from both sides
divide both sides by 4
4. Solve for x: e x+5 = 7∙2 x
e x+5 = 7∙2 x
ln (e x+5 ) = ln (7∙2 x )
(x+5)ln e = ln 7 + x ln 2
x+5 = ln 7 + x ln 2
x – x ln 2 = ln 7 – 5
x(1 – ln 2) = ln 7 – 5
x= (ln 7 – 5)/ (1 – ln 2)
take ln of both sides
use property: ln (ab) = ln a + ln b and ln ax = x ln a
since ln e = 1
group x terms: subtract 5 and x ln 2 from both sides
factor out x
divide both sides by (1 – ln 2)
4
5. Let p = ln m and q = ln n . Write in terms of p and q using no logs: ln (n m )
ln (n m4) = ln n + ln m4
ln (n m4) = ln n + 4 ln m
= q + 4p
use property: ln (ab) = ln a + ln b
use property: ln ax = x ln a
6. The number of asthma sufferers in the world, N (in millions) can be modeled by the function N = 84e
where t is the number of years after 1990.
0.0397 t
a. How many asthma sufferers were there in 1990?
t=0
N = 84 e
0.0397·0
= 84
N = a bt
b. Re-write this model in the form
b=e
0.0397
= 1.0405 N = 84 (1.0405)
.
t
c. Using either form of the equation, determine how long would it take for the initial number of asthma
sufferers to double.
There will be 168 million sufferers when doubled, so
t
84 (1.0405) = 168
t
(1.0405) = 2
t log 1.0405 = 2
t = 2/( log 1.0405) = 17.459
d. Check your answer by using the original equation or (b).
84 (1.0405) 17.459 = 168
So when t=17.459, N=168.
7. The concentration C (in milligrams per liter) of theophylline (a common asthma drug) in the blood stream
after an initial injection of 300 mg can be modeled by the function C = 11.9(0.84) t , where t is time
measured in hours.
a. Re-write this model in the form C = ae .
kt
k
e = 0.84 so k = ln (0.84) = -0.17435
k t
-0.17435 t
C = 11.9 (e ) = 11.9 e
b. What is the initial concentration in the blood stream?
When t = 0, C = 11.9
,
c. How long will it take for there to be half the initial concentration? Check your answer.
Use either equation:
t
(0.84) = ½
t = log 0.5 / log 0.84 = 3.98
Check the answer:
3.98
11.9 (0.84)
= 5.945
d. The value obtained in part (c), the half-life, is the time it takes for a decaying quantity to decrease by a
factor of 2 (by 50%). Using this value, determine how long it will take for there to be one 1/8th of the
original concentration. Hint: you do not need to solve an exponential equation to find this!
After 4 hours, we get that it is ½ the initial concentration.
So, after another 4 hours, after 8 hours total, it is ½ of that ½: ½ · ½ = ¼
Then, after 12 hours, it is ½ more: ½ · ½ · ½ = 1/8
8. In a 1994 episode of Seinfeld, Jerry’s Uncle Leo was supposed to give Jerry’s mother $50 that their father
had won at the racetrack in 1941. Nana says to Leo, “your father won $1000 at the track last week and he
gave you $100, and you were supposed to give $50 to your sister”. When Jerry’s father, Morty, finds out, he
says: “Do you know what the interest on that $50 comes to over 53 years? $663.40, and that’s figuring
conservatively at 5% interest over 53 years, compounded quarterly”.
a. Is Morty correct? How much interest would be in the account after 53 years?
$50 compounded quarterly at 5% interest is computed by:
50 (1 + 0.05/4) 4·53 = 50 (1.0125) 4·53 = 50 (1.0125) 212 = 696.17
Since $696.17 is the total amount, including the original $50, the interest is $696.17 – 50 = $646.17
Morty was incorrect, but close.
b.
If Uncle Leo agrees to put $200 in a trust fund for Jerry giving 5% interest compounded continuously,
how long would it take for the account to have $663?
Putting $200 in the bank at 5% interest compounded continuously, looking for a result of 663
is computed by:
0.05 t
200 e
= 663
0.05 t
e
= 663/200
0.05 t
e
= 3.315
ln e 0.05 t = ln 3.315
0.05 t = ln 3.315
0.05 t = 1.198476
t = 1.198476/0.05 = 23.969
So it would take 23.969 years to make $663 at 5% compounded continuously.