1. evaluate the exponential equation for three positive values of x, three negative values
of x, and at x=0. Show work. Use the resulting ordered pairs to plot the graph.
State the equation of the line asymptotic to the graph (if any). y=-(1/5)^x.
x
y = -(1/5)x
(x,y)
3
2
1
0
-1
-2
-3
-(1/5)3 = -(1/53) = -1/125
-(1/5)2 = -(1/52) = -1/25
-(1/5)1 = -(1/51) = -1/5
-(1/5)0 = -(1/50) = -1
-(1/5)-1 = -(1/5-1) = -5
-(1/5)-2 = -(1/5-2) = -25
-(1/5)-3 = -(1/5-3) = -125
(3, -1/125)
(2, -1/25)
(1, -1/5)
(0, -1)
(-1, -5)
(-2, -25)
(-3, -125)
The asymptote is y = 0.
The graph looks like this:
Use the formula N=Ie^kt, where N is the number of items in terms of the initial
population I, at time t, and k is the growth constant equal to the percent of growth per unit
of time. A certain radioactive isotope has a half-life of approximately 1100 years. How
many years would be required for a given amount of this isotope to decay to 40% of that
amount?
t = t1/2 ( ln(N/N0) / ln(0.5) ) = 1100( ln 0.4 / ln 0.5 ) = 1454.12 years
a. 1414 years
b. 811 years
c. 1454 years
d. 660 years
3. express as a sum, difference, and product of logarithms, without using exponents.
logb fourth root x^5b^9/y^7z^10
logb(4√x5b9/y7z10) = logb(4√x5b9) – logb(4√y7z10)
= logb(x5/4) + logb(b9/4) – logb(y7/4) – logb(z10/4)
= (5/4) logb x + 9/4 – (7/4) logb(y7) – (10/4)logb(z)
= (5/4) logb x + 9/4 – (7/4) logb(y7) – (5/2)logb(z)
a.5/4logb x + 9 - 7logb y - 10 logb z
b. 5/4logb x + 9/4 - 7/4 logb y -5/2logb z
c. 1/4(logb x^5 + logb b^9 -logb y^7 + logb z^10)
d. 5/4logb x + 9/4 - 7/4 logb y + 5/2logb z
4. choose the expression or equation that is equivalent to the one given. 1n x=2
a. x^e=2
b. e^x=2
c. x=1n 2
d. x=e^2
c. -6
d. 2
c. 3233loga a
d. 3233.
5. simplify log6 1/36
log6 (1/36) = log6 (1/62) = log6 (6-2) = -2
a. 6
b. -2
6. simplify loga a^3233
loga a3233 = 3233 loga a = 3233(1) = 3233
a. a3233 o
b. 1