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Name:
Grade:
out of 20
Intro to Calculus - Summer 2009
Instructor - Mike Limarzi
Daily Quiz 10 - Thursday, July 16, 2009
1. (5 points)
Rewrite as a sum or difference of multiples of logarithms.
√
3 xy
.
ln
t4/3
√
3 xy
√
= ln 3 xy − ln(t4/3 )
ln
t4/3
4
4
= ln (xy)1/3 − ln t = ln(x)1/3 + ln(y)1/3 − ln t
3
3
1
1
4
= ln x + ln y − ln t .
3
3
3
2. (5 points) Rewrite as a single expression.
3 log4 (x2 ) − 4 log4 (x−3 ) + 2 log4 (x).
3 log4 (x2 ) − 4 log4 (x−3 ) + 2 log4 (x) = log4 (x6 ) + log4 (x12 ) + log4 (x2 ).
= log4 (x20 ) .
3. (5 points) Solve the following equation.
log3 (x) = log3 2 − log3 (x − 2).
2
log3 (x) = log3
. Now raise 3 to the both sides:
x−2
2
x = x−2
. We cross-multiply to get:
2
x − 2x = 2. Now, we bring the 2 over and use the quadratic formula:
x2 − 2x − 2 = 0.
√
√
√
2± 4+8
2±2 3
x=
=
= 1 ± 3.
2
2
√
But, we need x > 0 and x − 2 > 0. Therefore, 1 − 3 is not a valid solution. This means we
√
have only one answer: 1 + 3 .
4. (5 points) Solve the following equation.
2x−1 3x+2
1
1
=
.
2
4
Take log2 of both sides:
2x−1
3x+2
1
1
log2
= log2
. Use the ”slap-it-over” rule:
2
4
1
1
(2x − 1) log2
= (3x + 2) log2
. Rewrite 12 = 2−1 and
2
4
1
4
= 2−2 .
(2x − 1) log2 (2−1 ) = (3x + 2) log2 (2−2 ). By properties of logs, we get:
(2x − 1)(−1) = (3x + 2)(−2). Simplifying:
−2x + 1 = −6x − 4. Solving:
4x = −5. So, our solution is x = −
Extra Credit: Name a type of cheese.
5
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4