Download Handout 4 Solutions Math 115 Spring 2011 1. Let h(x) = log 3(x − 5

Handout 4 Solutions Math 115 Spring 2011
1. Let h(x) = log3 (x − 5) + 4
(a) State the domain and range of h(x)
Solution: To find the domain, recall the input of a logarithm (of any base) must
be positive. So, x − 5 > 0, so x > 5. Thus, dom(h(x)) = {x : x > 5} . The
range of the toolkit logarithm function is all real numbers; we have shifted right
5 and up 4, but adding 4 to ‘all real numbers’ does not change the set. Thus,
ran(h(x)) = {x : x ∈ R} .
(b) find an equation for h−1 (x)
solution: To find the equation for an inverse, we ‘switch y and x’ and solve for
y.
x = log3 (y − 5) + 4
x − 4 = log3 (y − 5)
3x−4 = y − 5
3x−4 + 5 = y
So, h−1 (x) = 3x−4 + 5
(c) State the domain and range of h−1 (x)
Solution: We know that the domain and range switch for a function and its
inverse, so we can use our sets from part (a). Thus dom(h−1 ) = {x : x ∈ R} and
ran(h−1 ) = {y : y > 5}
(d) Carefully sketch the graph of y = h(h−1 (x))
Solution: By definition of inverse functions, h(h−1 (x)) = x for all x in the domain of h−1 (x). Thus we should plot y = x on the domain of the inverse, which
we found in part (c) to be all real numbers. The graph is shown below:
1
Figure 1: y = h(h−1 (x))
2. The present concentration of CO2 in the atmosphere is 379 ppm. If the concentration
increases at a yearly rate of .4%, how long will it take for the concentration to reach
500 ppm?
Solution: To find out how long it will take for the concentration to reach 500 ppm,
it would be very helpful to have an equation which gives concentration as a function of
time. We want a function of the form f (t) = abt where a is the initial amount, b is the
growth factor, and t is in years. We know the initial amount is 379ppm, so a = 379.
The yearly growth rate is .4%, so to convert to growth factor we convert to decimal
and add 1. The growth factor is then b = 1.0041 . Now that we have a function, we
can solve the equation f (t) = 500 for t:
500 = 379(1.004)t
500
= 1.004t
379
500
ln
= ln(1.004t )
379
500
= t ln(1.004)
ln
379
500
ln( 379
)
t=
ln(1.004)
So, the concentration will reach 500ppm after approximately 69.406 years .
3. Solve the following equations algebraically. Give exact answers.
1
Make sure to convert .4% to decimal correctly!
2
(a) (2.1)ex/2 = 5
(2.1)ex/2 = 5
5
ex/2 =
2.1
5
ln(e ) = ln
2.1
5
x
= ln
2
2.1
x/2
5
So, x = 2 ln( 2.1
)
(b) log(x) + log(x + 1) = log(3)
log(x) + log(x + 1) = log(3)
log(x(x + 1)) = log(3)
10log(x(x+1)) = 10log(3)
x(x + 1) = 3
x2 + x − 3 = 0
√
−1 ± 13
x=
2
One of these solutions is negative, so we must throw it away. Thus x =
√
−1+ 13
2
(c) 2 ln(x) = ln(3)
2 ln(x) = ln(3)
ln(x2 ) = ln(3)
2
eln(x ) = eln(3)
x2 = 3
√
x=± 3
Again, we have to throw away the negative solution because we can’t take the
√
logarithm of a negative number. Thus x = 3 .
3
(d) ln(2x + 1) − 1 = ln(x − 2)
ln(2x + 1) − 1 = ln(x − 2)
ln(2x + 1) − ln(x − 2) = 1
2x + 1
ln
=1
x−2
2x+1
eln( x−2 ) = e
2x + 1
=e
x−2
2x + 1 = e(x − 2)
1 + 2e = x(e − 2)
And thus x =
(e)
log(x+1)
log(x−1)
1+2e
e−2
=2
log(x + 1)
=2
log(x − 1)
log(x + 1) = 2 log(x − 1)
2
10log(x+1) = 10log((x−1) ) x + 1
x2 − 3x = 0
x(x − 3) = 0
= (x − 1)2
We thus have x = 0 or x = 3, but, x = 0 leads to a negative in a log, so we must
eliminate it. Hence x = 3
4. The ‘doubling time’ for a bank account balance that earns compound interest is given
by:
ln(2)
N= (1)
k ln 1 + kI
where N is the doubling time in years, I is the annual interest rate (in decimal form),
and k is the number of compounding periods per year.
(a) How long does it take for an investment to double at 5% interest compounded
monthly?
Solution: We can simply evaluate the formula for I = .05 and k = 12:
ln(2)
N=
ln
1+
.05 12
12
4
≈ 13.8918 years
(2)
(b) What annual interest rate, compounded annually, will cause your money to double
every 10 years?
Solution: This time, we know that k = 1 and N = 10, and we need to solve for
I:
ln(2)
ln ((1 + I))
10 ln(1 + I) = ln(2)
ln(2)
ln(1 + I) =
10
ln(2)
1 + I = e 10
10 =
I=e
≈ 0.0717
So, I ≈ 7.17% .
5
ln(2)
10
−1