Chapter 1 – Worksheet 4 - LOGARITHMS NAME________________________________

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Chapter 1 – Worksheet 4 - LOGARITHMS
NAME________________________________
1. Calculate the following. Whenever possible give the exact value. If the value does not exist, state DNE.
A. log 1.035
0.01494
 1 
B. log 4  
 64 
-3
C. log 2  4 
DNE
D. ln  e 
E. log9  3

½
On some of these work should be shown.
2. a) Using the change of base formula, write y  log a  x  using log base 10.
y  log a ( x) 
log( x)
log( a)
b) Using the change of base formula, write y  log a  x  using log base e.
y  log a ( x) 
ln( x)
ln( a)
3. Derive the formula for the tripling time for P  P0 a nt . What does this tripling time NOT
depend on? Do we need to make any assumptions about a and n ? If so, what are the assumptions?
ln 3
Since P0 triples this means P  3P0  3P0  P0 a nt  3  a nt  ln 3  nt ln a  t 
n ln a
So tripling time does not depend on P0 , the initial amount
“n” and “a” are positive.
1
Considering the values of a  3 and n 
tripling  time
4. Consider S  D   0.159  0.118log  D  . S is the slope of a beach and D is the average diameter
(in mm) of the sand particles on the beach. Suppose a particular beach rises 9 meters for every 100
Meters inland. What size sand would you expect to find on that beach?
9
9
 0.159
 0.159
100
9
100
0.118
 0.159  0.118 log D  log D 
 D  10
 .26 mm.
100
0.118
So the sand size is approximately 0.26 mm
5. You and a friend plan to purchase cars in September. The initial value of your car will be $34,000
and will depreciate 17% each year. The initial value of your friend’s car will be $16,500 and will
depreciate 12% each year. You agree to exchange cars when their values are equal.
A. How long do you need to wait? (to the nearest month) What is the value of your car?
V1 (t )  34,000(0.83) t and V2 (t )  16,500(0.88) t so we set these equal and solve for time.
 34000 
ln 

t
34,000  .88 
16500 

t
t
34,000(0.83 )  16,500(0.88 ) 

 12.36 years
 t 
16,500  .83 
 0.88 
ln 

 0.83 
So in about 12.36 years their cars are equal in value. Value is $3398.48
B. What would your depreciation rate have to be in order for the values of the cars to match at the
end of 7 years? (assume your friend’s car depreciates 12% each year)
7
1
16,500
b
 16,500  7
 b 
34,000(b 7 )  16,500(0.88 7 )  


 
 
34,000
0.88  34,000 
 0.88 
1
 16,500  7
b  0.88
  0.7936  r  1  0.7936  0.2064
 34,000 
So your depreciation rate would have to be about 20.64% annually for the cars to be equal in value
in seven years.
6. A. Sketch a graph of y  7
log7 x
. Consider and include the domain in your sketch.
Domain is x  0 . Graph looks like y  x in first quadrant. Open circle at (0,0)
B. Sketch a graph of y  ln e x . Consider and include the domain in your sketch.
Domain is all reals. Graph looks like y  x.
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