College Algebra to Calculus and the TI-84
Lesson #14
Logarithmic functions
Exercise 1. Graph the function y  log 3 x



WINDOW Xmin=-3 Xmax=30 Ymin=-10 Ymax=5
Y= (deselect any existing functions and plots)
Y1=LOG(x)LOG(3)
GRAPH
Exercise 2. Graph the function y=LN(x)
This is called the logarithmic function.

Y=
Y1= LN(X)
GRAPH
Exercise 3. Evaluate log 3.78 2.91 to the nearest 5 decimal places.
 MODE (select 5 decimal places)
 LOG(2.91)  LOG(3.78) ENTER or
LN(2.91)  LN(3.78) ENTER
answer: 0.80329
Exercise 4. Evaluate Invlog(-3.0573) to 6 decimal places.
 MODE (select 6 decimal places)
 2nd LOG (-3.0573)
ENTER
answer: 0.000876
7.385.3
5
93.8
LN (5.3  LN(7.38) - LN(93.8)  5) ENTER
Exercise 5. Use LN to compute

2nd
answer: 16078.96723
Exercise 6. Solve the logarithmic equation Log(x-3) + log(x+2) = Log 14


MATH 0 ▲ CLEAR eqn: LOG(X-3) + LOG(X+2) - LOG (14) ENTER
x= 10 (or any other guess) ALPHA SOLVE
answer: 5
Exercise 7. Solve the equation log x 3  6
 MATH 0 ▲ CLEAR
eqn: 0 = LN(3)  LN(X) - 6
 x= 5 (any guess)
ALPHA SOLVE
answer: 1.20093695518
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ENTER
Exercise 8. Solve the equation log 4 x  2 , give answer as a fraction.
 MATH 0 ▲ CLEAR eqn: 0=LN(X)  LN(4) + 2 ENTER
 x=1.20009
ALPHA SOLVE
answer: 0.06249999999
nd
 2 QUIT
X ENTER
 MATH 1 ENTER
answer: 1/16
Exercise 9. Graph the function f(x) = log(2x-8) and its inverse.
 WINDOW Xmin = -1
Xmax = 10 Ymin =- 2 Ymax = 10
 Y= Y1= LOG(2x - 8) GRAPH
 2nd DRAW 8 VARS Y-VARS 1 1 (to select Y1)
 2nd DRAW 1
ENTER
Exercise 10. Find the approximate solution of the equation 2.53 -xlog(34) = log 76.
Round answer to four decimal places.
 MODE (select four decimal places)
 MATH 0 ▲ CLEAR eqn: 0= 2.53 -xLOG(34) - LOG 76 ENTER
 x = 1 (or any guess) ALPHA SOLVE
answer: .4239
Exercise 11. Find the approximate two solutions of the equation x-2 = ln(x). Round
answer to four decimal places
 MATH 0 ▲ CLEAR eqn: 0 = x - 2 - LN(x) ENTER
 x = 0 .4 (one guess) ALPHA SOLVE
answer: x = 0.1586
 x = 5 (second guess) ALPHA SOLVE
answer: x = 3.1462
Exercise 12. Consider the function y = xLn(x) for values 0<x<2. Note: x>0
a) Graph the function
 Y=
y1=xLN(x)
 WINDOW Xmin = -0.2 Xmax = 2 Ymin=-1 Ymax = 2 GRAPH
b) Find the x-intercept of the function
 GRAPH 2nd CALC 2 Left Bound? ENTER
 Guess? ENTER
answer: x = 1
Right Bound? ENTER
c) Find the minimum value of the function.
 GRAPH 2nd CALC 3 Left Bound? ENTER Right Bound? ENTER
 Guess? ENTER
answer: minimum value is y = -0.3678794 at x=0.36787819
or 2nd QUIT
 MATH 6 fMin(xLn(x), x, 0, 2) ENTER answer: x= 0 .3678789354
 VAR Y-VAR 1 1
 Y1(2nd ANS) ENTER
answer: y = -0.3678794412
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Exercise 12. Continuation
d) Draw the tangent line at x = 1 and find the equation of the tangent line.
 GRAPH 2nd DRAW 5
1 ENTER answer: y=1x – 1 (approximately)
b) Draw the tangent line at x=.5 and find the equation of the tangent line.
 GRAPH 2nd DRAW 5
0.5
ENTER answer: y=0.31x -0.5
(approximately)
 2nd DRAW 1 2nd QUIT
f) Shade the area between the x-axis and the curve from x=0 to x=1.5
 GRAPH 2nd DRAW 7 Shade( xLNx, 0, 0, 1, 1, 2) ENTER
 2nd DRAW 7 Shade(0, xLNx, 1, 1.5, 1, 2) ENTER
 2nd DRAW 1
2nd QUIT
g) Find the solution of the equation xLn(x) = e  x
 Y=
y2=e^(-x) + GRAPH
nd
 2 CALC 5 First curve? ENTER
 Second curve? ENTER
 Guess? Trace to a point close to the intersection ENTER
Answer: x=1.2550137
y=.28507195
The solution of the equation is x=1.2550137
or
 MATH 0 ▲ CLEAR 0=xLN(x)=e^(-x) ENTER
 X=1 (any guess) ALPHA SOLVE 1.2550136915
Exercise 13. A car depreciates exponentially at a 20 % annual rate according to the
formula y(t )  14000(.80) t ,where y is the value of the car and t is the number of years.
a) Store and graph the formula as y1 and use 2nd TABLE to find y(0), y(1 year), y(15
months), y(4 years)
 Y=
y1=14000(.80)^x
 WINDOW Xmin=-1
Xmax=10
Ymin=-1000
Ymax=16000 GRAPH
nd
 2 TBLSET
TblStart=0
Tbl  0.25
nd
 2 TABLE y(0)=14000, y(1) =11200 y=(1.25) =10592 y(4) =5734.4
b) Use logs to find in how many years will the value of the car depreciate to $2000.
Ln( y1 )  Ln(14000)
t
Ln(.80)
 (LN(2000)-LN(14000))  LN(.80) answer: 8.72 years
Check:
 2nd TABLE y1(8.72)=2000
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Exercise14. Suppose $5000 is invested at an annual interest rate of 8%, compounded
monthly. In how many years will the original money of $5000 double.
P=$5000, B=$10000, r=8%, k=12, n = kt , B(t) = B, find t.
Note:
kt
kt


r
r 
r 

B
B
Formula: B(t )  P 1    ln    ln  1     ln    kt ln  1   
k
k 
k 

P
P



10000
B
ln
ln

r 
1
1
B
5000
P
 kt ln  1     ln    t  


k
P
r
k
12

 0.08 

 

ln  1  
ln  1 

12 
 k

 (1  12)  (LN(100005000))(LN(1+0.0812)) answer: 8.69
Check by using FINANCE
TI-84: APPS 1: FINANCE 1 TVM Solver
N=180 I%=8 P=-5000
PMT=0 FV=10000
 move back to N=180
ALPHA SOLVE
answer: 104.3182669
 104.3182669  12 ENTER
answer: 8.69 years
Exercise 15. In exercise 14, in how many years will the original money of $5000 double
if compounded continuously?
Formula:
B(t )
 B(t ) 
 B(t ) 
B(t )  Pe rt  e rt 
 Ln e rt  Ln 

rtLn
(
e
)

Ln

 P 
P
 P 


1  B(t ) 
 B(t ) 
 rt  Ln 
because
Ln
(
e
)

1

t

Ln

r  P 
 P 
P=5000, r=8%, B(t)=10000
 (1  .08)LN(10000  5000)
answer: 8.66 years
Check answer
 
TI-84: APPS 1: FINANCE 1 TVM Solver
 N=15  EE10 (or any large number) I%=8 PV=-5000 PMT=0
FV=10000 (or any other guess)
 P/Y=EE10 C/Y=EE10 PMT: END
 (go back to N) ALPHA SOLVE answer: 8.66E10 periods
answer: 8.66 years
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