Name
April 1, 2014
Honors Math 3 Sample Test
Chapter 8: Logarithmic Functions page 1
Instructions: Calculators are allowed on this test. Show all work for credit
1. Write below the three Laws of Logarithms.
2. Prove the Third Law by using the corresponding Law of Exponents and the Magic
Wand.
3. Find the value of 3log3 75 . Show your work.
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Name
April 1, 2014
Honors Math 3 Sample Test
Chapter 8: Logarithmic Functions page 2
4. Solve the following equation.
log(x + 3) + log(x + 5) = log(8)
5. Let a and b stand for positive numbers greater than 1.
How are the quantities logb(a) and loga( 1b ) related to each other?
Prove your answer.
6. Solve the following equation. Be sure to check whether your solutions are valid.
log x + log( x - 3) =1
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Name
April 1, 2014
Honors Math 3 Sample Test
Chapter 8: Logarithmic Functions page 3
7. Answer this question by setting up an equation and solving it.
a. Consider $500 invested at a 6% rate per annum, compounded monthly. This means
that each month, interest is paid on the account at the rate of 6% ÷ 12 = ½ %.
Write an exponential equation for the amount in the account, A(t), as a function of the
number of months t from when the account was opened.
b. Solve your equation for t to find how many months will it take for the value of the
investment to double (reach $1000).
SHOW YOUR WORK
Name
April 1, 2014
Honors Math 3 Sample Test
Chapter 8: Logarithmic Functions page 4
ANSWERS
Please read the following answers carefully, noting that I will be expecting you to write
REASONS for each step.
1. Write below the three Laws of Logarithms.
1. log b M + log b N = log b ( M × N )
2. log b M - log b N = log b ( M / N )
3. log b M N = N × log b ( M )
2. Prove the Third Law by using the corresponding Law of Exponents and the Magic
Wand.
1. Let x = logb M Û M = b x .
( )
2. Then M N = b x
N
(Magic Wand)
= b x×N = b N×x .
3. Û logb M N = N × x = N × logb M
(3rd Law of Exponents)
(Magic Wand)
3. Find the value of 3log3 75 . Show your work.
3log3 75 = 75, because f (x) = 3x and g(x) = log3 x are inverse functions.
4. Solve the following equation.
log(x + 3) + log(x + 5) = log(8)
log[(x + 3)(x + 5)] = log(8)
(1st Law of Logarithms)
(x + 3)(x + 5) = 8
(Logarithms are one-to-one)
x 2 + 8x +15 = 8
x 2 + 8x + 7 = 0
(x + 7)(x +1) = 0
x = { –7, –1}
Plugging these values into the original equation, we have
log(–7 + 3) + log(–7 + 5) = log(–4) + log(–2) is undefined.
log(–1 + 3) + log(–1 + 5) = log(2) + log(4) = log(8) √
Thus –7 is extraneous and –1 is a solution.
SHOW YOUR WORK
Name
April 1, 2014
Honors Math 3 Sample Test
Chapter 8: Logarithmic Functions page 5
5. Let a and b stand for positive numbers greater than 1.
How are the quantities logb(a) and loga( 1b ) related to each other?
Prove your answer.
Here are two different ways to solve this:
Let x = logb a Û a = b x .
Then b = a x and
1
1
1
= a- x .
b
(Magic Wand)
(3rd and 2nd Laws of Exponents)
æ1ö
1
1
Using the Magic Wand, we get log a ç ÷ = - = èbø
x
log b a
Second way:
Use the Base Change Formula log b y =
log b a =
log a y
with y = a:
log a b
log a a
1
, since loga a =1.
=
loga b loga b
æ1ö
But loga ç ÷ = loga (1) - log a b = -loga b ,
èbø
1
-1
so logb a =
, as before.
=
log a b log a (1/ b)
(2nd Law of Logarithms)
6. Solve the following equation. Be sure to check whether your solutions are valid.
log x + log( x - 3) =1
log[x ( x - 3)] =1
(1st Law of Logarithms, then Magic Wand)
1
x(x - 3) =10
x 2 - 3x -10 = 0
(x - 5)(x + 2) = 0
x = { 5, –2}
Plugging these values into the original equation, we have
log (5) + log (5 – 3) = log (5) + log (2) = log (10) = 1 √
log(–2) + log(–2 – 3) = log(–2) + log(–5) is undefined
Thus –2 is extraneous and 5 is a solution.
SHOW YOUR WORK
Name
April 1, 2014
Honors Math 3 Sample Test
Chapter 8: Logarithmic Functions page 6
7. Answer this question by setting up an equation and solving it.
a. Consider $500 invested at a 6% rate per annum, compounded monthly. This means
that each month, interest is paid on the account at the rate of 6% ÷ 12 = ½ %.
Write an exponential equation for the amount in the account, A(t), as a function of the
number of months t from when the account was opened.
A(t) = 500 × (1.005)t
(Exponential function with G.F. = 1 + ½ %, t in months)
b. Solve your equation for t to find how many months will it take for the value of the
investment to double (reach $1000).
A(t) = 1000 = 500 × (1.005)t
1000
= 2 = (1.005)t
500
log 2
t=
@ 139 months.
log1.005
The last step uses the Magic Wand and the Base Change Formula.
SHOW YOUR WORK