MATH 150
Quiz Key #8
11/10-12/2015
(1) Solve the equation ln 3 + ln (2x + 9) = 2 ln x.
Solution:
Use properties of logarithms to simplify both sides as follows:
ln [3 (2x + 9)] = ln x2 .
3 (2x + 9) = x2 .
x2 − 6x − 27 = 0.
( x − 9) ( x + 3) = 0.
x = 9 or x = −3.
Now we need to verify which solutions actually work in the original equation.
If we plug in x = 9 to the left-hand side, we get: ln 3 + ln (27) = ln (81) . If
we plug into the right-hand side, we get: 2 ln (9) = ln (81) . So this solution
works. Now, we check the other. Plugging x = −3 into the left-hand side,
we get ln 3 + ln 3 = ln 9. Plugging into the right-hand side we get 2 ln (−3) .
However, we cannot take the logarithm of a negative number (when working
with real numbers). Therefore, our only solution is x = 9.
(2) Solve the equation 5e2x + 13e x − 6 = 0.
Solution: Note that e2x = (e x )2 . Therefore, we can factor the equation as
(5e x − 2) (e x + 3) = 0.
2
or e x = −3.
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Now the range of y = e x is (0, ∞) , so e x is never equal to −3. Therefore,
ex =
2
.
5
2
x = ln
.
5
ex =
1
(3) Answer the following equations for the function f ( x ) = −3 log2 ( x + 3) +
4.
Solution: The basic function is y = log2 x. We then reflect about the y-axis and
stretch vertically by a factor of 3. Following this, we shift left 3 and up 4. The
domain of the original function is (0, ∞) . The only transformation affecting the
domain is the shift left 3. Therefore, the DOMAIN is (−3, ∞) . The range of the
original function is (−∞, ∞) . Reflecting this, stretching it, and shifting it up 4
gives the RANGE as (−∞, ∞) . To find the x-intercept, we solve the following
equation:
0 = −3 log2 ( x + 3) + 4.
4
= log2 ( x + 3) .
3
24/3 = x + 3.
x = 24/3 − 3.
Therefore the x-INTERCEPT is 24/3 − 3, 0 . The y-intercept is found by finding f (0) = −3 log2 3 + 4. Therefore, the y-INTERCEPT is (0, −3 log2 3 + 4) .
Finally, the base function is increasing. Therefore, when we reflect about the
y-axis, the new function is DECREASING on its domain.
(4) If ln 2 = a, ln 3 = b, ln 5 = c, and ln 11 = d, evaluate and fully simplify
log11 25
36 .
Solution:
log11
25
36
ln 25
ln 25 − ln 36
2 ln 5 − 2 ln 6
2 ln 5 − 2 ln 2 − 2 ln 3
2c − 2a − 2b
36
=
=
=
=
=
.
ln 11
ln 11
ln 11
ln 11
d
(5) Fully simplify 2 log10 6 − 2 log10 15 − 4 log10 2.
Solution:
2 log10 6 − 2 log10 15 − 4 log10 2 = log10 36 − log10 152 − log10 16
= log10
22 · 32
32 · 52 · 24
= log10
1
100
= − log10 100 = −2.
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(6) What is the half-life of a sample that decayed 39% after 4 years? Include
the units in your answer.
Solution: 61% of the sample remains after 4 years. The decay equation is
A (t) = A (0) · ekt . Therefore,
A (4) = .61A (0) .
A (0) e4k = .61A (0) .
e4k = .61.
4k = ln (.61) .
k=
ln (.61)
.
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We want to find t such that A (t) = 12 A (0) . Therefore,
A (0) ekt =
1
A (0) .
2
1
ekt = .
2
1
kt = ln
.
2
ln 12
4 ln 12
t=
=
years.
k
ln (0.61)
(7) Suppose that a population of mice with initial population P triples every
20 days. What is the exponential growth model where t is time in days?
Solution: P (0) = P. P (20) = 3P. P (40) = 3 (3P) = 32 P. P (t) = P · 3t/20 .
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(8) How many mL of 30% saline and how many mL of 55% saline should
be mixed to get 100 mL of a 40% saline solution? Include the units in your
answer.
Solution: Let x be the number of mL of 30% saline and y be the number of mL
of 55% saline. Then we get the following system of equations:
30x + 55y = 40 (100)
.
x + y = 100
Isolating y in the second equation yields y = 100 − x. Substituting back in, we
get
30x + 55 (100 − x ) = 4000.
30x + 5500 − 55x = 4000.
−25x = −1500.
1500
= 60.
x=
25
Therefore, y = 100 − 60 = 40. Therefore, we should mix 60 mL of 30% saline
solution and 40 mL of 55% saline solution.
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