Solutions to LOG Homework

 

1.    a)  log 6 216 = 3                           b) 10 0.59 = 3.9                           c)  8 0.88 = 6.2

2.    a) 7.2 log 9 81 = (7.2) log 9 9 2 = (7.2) 2 log 9 9 = (7.2)(2)(1) = 14.4

       b) log 400 = log 4 • 100 = log 4 + log 100 = 0.6021 + 2 = 2.6021

       c) 4 log 7 49 = 4 log 7 72 = 4 2 log 7 7 = 4 2(1) = 16

3.    a) 9.1 log 3 5 = 9.1(1.4650)= 13.33                 

       b) log 3 18 = log 3 3•6 = log 3 3 + log 3 6 = 1 + 1.6309 = 2.63

       c)7 log 3 5 = 7 1.4650 = 17.3

4.    a)   2.1 + 0.6 e 5 x + 2 = 10             (Subtract 2.1 from both sides)

          – 2.1                       – 2.1

                     0.6 e 5 x + 2  = 7.9            (Divide both sides by 0.6)

                  

                        e 5 x + 2 = 13.1666667    (Now take LN of both sides)

 

                  LN(e 5 x + 2)  = LN(13.1666667)

 

                     5 x + 2 =  2.577688

 

                     5 x = 0.57688

 

                        x = 0.11537 or 0.12

 

 

      b)              12 log(3 + 2x) = 72               (Divide both sides by 12.)

 

                          log(3 + 2x) = 6                   (Now raise 10 to the indicated powers on both sides)

 

                  

                     10 log(3 + 2x) = 10 6

 

                     3 + 2x = 1,000,000

 

                        2x  =   999,997

 

                          x =  499998.5

 

      c)     4(10 3x + 1) – 8 = 8                       (Add 8 to both sides)

 

              4(10 3x + 1)   =  16                       (Divide both sides by 4)

 

                (10 3x + 1)   = 4                         (Take the LOG of both sides)

 

         LOG(10 3x + 1)  =  LOG(4)

 

                3x + 1  =  0.60205

 

                    3x  = - 0.39794

 

                     x = - 0.13263 or -0.13

 

 

d)          9.1 + ln(4x – 1) = 6.8                   (Subtract 9.1 from both sides)

 

           ln(4x – 1) = – 2.3                      (Now raise e to the indicated powers on both sides)

 

       e ln(4x – 1) =  e – 2.3

 

         4x – 1 = 0.1002588

 

            4x =  1.1002588

 

              x =  0.27506   or 0.28

 

5. Applications of log functions.

 

a)             L(x) = 3 log x, where x is the weight of the plane in thousands of pounds. So, the Puddle-Jumper plane is 235 thousands of pounds.

   

Therefore, I need to know how long a runway, in thousands of feet, is needed to land/take off safely for this plane.

 

L(235) = 3 log (235)

 

           = 7.113

 

This value means that the minimum runway length is 7, 113 feet for safe take off/landing of the Puddle-Jumper plane. So, 7,500 feet is safe for landing the plane.

 

 

 

b)            P(x) = 0.48 ln(x + 1) + 27 models the barometric pressure of a hurricane, where x is the number of miles from the eye of the storm.

I want to know the pressure when I am 2.3 miles from the storm.

 

P(2.3) = 0.48 ln(2.3 +1) + 27

 

          = 0.48 ln(3.3) + 27              (Get ln(3.3) from the calculator, multiply by 0.48, then add 27)

 

          = 0.48(1.193922) + 27

 

          =  27.5730  or 27.57 is the barometric pressure when 2.3 miles from the eye of a hurricane