Math 133 Unit 5 Essay
Student Answer Form Unit 5 IP_2
April 22, 2011
(1.)
a.
e^(0.05t)=1600
Answer: t=147.556
Show your work:
ln(e^(0.05t))=ln(1600)
0.05t=ln(1600)
(0.05t)/(0.05)=(ln(1600))/(0.05) t=(ln(1600))/(0.05) t=20ln(1600) t=147.556 b.
ln(4x)=3
Answer: x= 5.0214
Show your work:
4x = e^3/4
c.
log2(8-6x) = 5
Answer: x= -4
Show work:
8 – 6x = 2^5
6x = 8 – 32
X = - 24/6
d.
4 + 5e – x = 0
Answer: no solution found
Show work: e ^ - x = - 4/5
(2.)
a.
g(x) = log (x + 5)
Description of transformation: shifted left 5 units
Equation for vertical asymptote: x = -5
X-intercept in (x, y) form: (- 4, 0)
b.
g(x) = log ( - x)
Description of transformation: no units were moved, stayed
April 22, 2011
(1.)
a.
e^(0.05t)=1600
Answer: t=147.556
Show your work:
ln(e^(0.05t))=ln(1600)
0.05t=ln(1600)
(0.05t)/(0.05)=(ln(1600))/(0.05) t=(ln(1600))/(0.05) t=20ln(1600) t=147.556 b.
ln(4x)=3
Answer: x= 5.0214
Show your work:
4x = e^3/4
c.
log2(8-6x) = 5
Answer: x= -4
Show work:
8 – 6x = 2^5
6x = 8 – 32
X = - 24/6
d.
4 + 5e – x = 0
Answer: no solution found
Show work: e ^ - x = - 4/5
(2.)
a.
g(x) = log (x + 5)
Description of transformation: shifted left 5 units
Equation for vertical asymptote: x = -5
X-intercept in (x, y) form: (- 4, 0)
b.
g(x) = log ( - x)
Description of transformation: no units were moved, stayed