Download Week 7 Stuph (and 6.1) 1. Solve: log5 (x − 2) + log 5 (2x − 9) = 2

Week 7 Stuph (and 6.1)
1. Solve:
log5 (x − 2) + log5 (2x − 9) = 2
log5 ((x − 2) (2x − 9)) = 2
52 = (x − 2( (2x − 9)
25 = 2x2 − 13x + 18
0 = 2x2 − 13x − 7
0 = (2x + 1) (x − 7)
1
x = − ,7
2
1
Now when we check our solutions, we see that x = − , doesn’t work, because when
2
we plug this value into our original equation, we are forced to take the log of a negative
number, which is not possible since there are no negative numbers in the domain of log.
x = 7, on the other hand, does not force us to take the log of a negative number, and works.
Therefore x = 7 is our only solution.
2. Solve:
35+4x = 9x
35+4x = 32x
5 + 4x = 2x
5 = −2x
5
= x
−
2
If we plug this value into our original equation, we would see that it checks out.
3. Solve:
e2x = 10.4
ln e2x = ln (10.4)
2x = ln (10.4)
ln (10.4)
x =
2
x ≈ 1.17
1
2
4. Suppose $5000 is invested at interest rate k, compounded continuously, and grows to
$6954.84 in 6 years.
(a) Find the interest rate k
Note that when t = 6, f (t) = 6954.84. Therefore we can conclude that
6954.84 = 5000ek·6
Now we can solve the above equation for k as follows:
6954.84
6954.84
5000
6954.84
ln
5000
ln 6954.84
5000
6
.055
= 5000ek·6
= e6k
= 6k
= k
≈ k
(b) Find the exponential growth function
f (t) = 5000e.055t
(c) Find the doubling time
We want to know when we have $10, 000, or double what we started with. So we have
to solve for t in the equation:
10, 000 = 5000e.055t
This is the same as solving for t in the following:
2 = e.055t
ln (2) = .055t
ln (2)
.055
= t
12.6 ≈ t
So it seems as if the money will double in approximately 12.6 years.
3
1
, where θ is an acute angle (ie, it
2
is not the right angle). Find the following (exact – not approximate):
√
3
(a) cos θ =
2
√
3
(b) tan θ =
3
√
2 3
(c) sec θ =
3
5. Suppose that there is a right triangle whose sin θ =
(d) csc θ = 2
√
(e) cot θ = 3
For the above problems, first we should make a triangle to help us out (if we haven’t
already memorized this triangle...)
1
Knowing that sin θ = , that tells us that the length of the side opposite θ is 1, and that
2
the length of the hypotenuse is 2. Using this information, and the pythagorean theorem,
we can find the length of the third side. So, we know that
12 + b2
1 + b2
b2
b
=
=
=
=
22
4
3
√
± 3
√
However since length is positive, the only measurement we deal with is positive 3. So
now that we know the length of each side of our triangle, we need only use the definition of
each trigonometric function to find the other ratios. Note the triangle below, which should
help (hopefully).