The document contains a 15 question chapter test on mathematics topics including: simplifying expressions, solving equations, applying logarithmic and exponential functions, compound interest, radioactive decay, and bacterial growth. The answers provide the step-by-step work to arrive at the numerical or algebraic solution for each question.

1. Chapter Test
1) Simplify each:
a) 4 . 2-4 = b) (4 . 2)-4 = c) (a-1 + b-1)-2 =
2) Simplify each:
3) Solve the equation: 93x = 81x + 1
4) The half life of a radioactive isotope is 7 days. If 5.6 kg are present now, how much will
be present in t days? In 13 days?
5) A bacteria colony triples every 6 days. The population currently is 5000 bacteria. What
will be the population in 21 days?
6) $1000 is invested at 8% compounded quarterly for 3 years. How much interest will you
receive at the end of the time period?
7) You invest $3000 at 7% compounded continuously for 2 years. How much money will
be in the account at the end of the time period?
8) Solve each log equation:
a) Log x = 21
b) Log |x| = 15
c) Ln (x2 - 1) = 3
d) Log x = 1.6 (Use calculator and round to hundredths.)
9) Write each log as a single log:
a) Log x + log y + 2Log z
b) Ln x + Ln y - 3Ln z
10) Simplify each log:

2. a) Ln e5 =
b) 102log4 =
c) 101 + log 5 =
11) Solve the equation: Log (x + 2) + Log 5 = 4
12) Graph y = 3x and y = log3 x on the same axis.
13) Use a calculator to find Log7 58 and round to hundredths place.
14) solve the equation: 15x + 1 = 29 and round to hundredths place.
15) How long will it take to double $1000 if invested at 6% compounded monthly?

3. Answers to the chapter test!!
1) Simplify each:
a) 4 . 2-4 = b) (4 . 2)-4 = c) (a-1 + b-1)-2 =
4(1/16) = 1/84 (1/a + 1/b)-2
= 1/4 = 1/4096 = [(a + b)/ab]-2
=[ab/(a + b)]2
= a2b2/(a + b)2
2) Simplify each:
a) (25/9)1/2 = 5/3 b) 161/2 = 4 c) x - 2
3) Solve the equation: 93x = 81x + 1
93x = 92(x + 1)
3x = 2(x + 1)
3x = 2x + 2
x = 2
4) The half life of a radioactive isotope is 7 days. If 5.6 kg are present now, how much will
be present in t days? In 13 days?
A(t) = 5.6(1/2)t/7
A(13) = 5.6(1/2)13/7 = 1.55 kg
5) A bacteria colony triples every 6 days. The population currently is 5000 bacteria. What
will be the population in 21 days?
A(21) = 5000(3)21/6 = 233827
6) $1000 is invested at 8% compounded quarterly for 3 years. How much interest will you
receive at the end of the time period?
A(3) = 1000(1 + .08/4)12 = $1268.24
7) You invest $3000 at 7% compounded continuously for 2 years. How much money will

4. be in the account at the end of the time period?
P(2) = 3000 e.14 = $3450.82
8) Solve each log equation:
a) Log x = 21
Write exponentially, 1021 = x
b) Log |x| = 15
Write exponentially and solve: | x | = 1015
x = 1015 or -1015
c) Ln (x2 - 1) = 3
Write exponentially and solve
e3 = x2 - 1
e3 + 1 = x2
_____
+ / e3 + 1 = x
d) Log x = 1.6 (Use calculator and round to hundredths.)
Write exponentially, 101.6 = x
x = 39.81
9) Write each log as a single log:
a) Log x + log y + 2Log z
=Log xyz2
b) Ln x + Ln y - 3Ln z
= Ln (xy/z3)
10) Simplify each log:
a) Ln e5 = 5
b) 102log4 = 16
c) 101 + log 5 = (10)(10log 5) = (10)5 = 50
11) Solve the equation: Log (x + 2) + Log 5 = 4
Log 5(x + 2) = 4

5. 5(x + 2) = 104
5x + 10 = 10000
5x = 9990
x = 1998
12) Graph y = 3x and y = log3 x on the same axis.
13) Use a calculator to find Log7 58 and round to hundredths place.
= log 58/log 7 = 2.09
14) Solve the equation: 15x + 1 = 29 and round to hundredths place.
(x + 1)log 15 = log 29
x log 15 + log 15 = log 29
xlog 15 = log 29 - log 15
x = (log 29 - log 15)/ log 15
x = .24
15) How long will it take to double $1000 if invested at 6% compounded monthly?
2000 = 1000(1 + .06/12)12t
2 = (1 + .06/12)12t
log 2 = 12t log (1 + .06/12)
log 2/ 12log(1 + .06/12) = t
11.58 = t
11.58 years