• 60 is a product: 60 = 2²×5×3. So log 60 = 2log 2 + log 5 + log 3. Notice that we used both the rule for products and the rule for powers here: log(x^y) = ylog(x).
• To evaluate log [2/55] we notice first that it's a quotient. The log of a quotient is the difference of the logs:
  

  log  2 55 = log 2 - log 55

  and we have a value for log 2. Now we see that 55 can be factored; and the log of a product is the sum of its factors' logs.
  

  log 2 - log 55 = log 2 - (log 5 + log 11)

  so that
  

  log  2 55 = 0.30103 - (0.69897+1.04139) = -1.42933

• Notice that 3500 is the product of 35 and 100, so we are going to use the rule for products here. We could work out all the factors of 100, but it's quicker to notice that 100 = 10² and so log(100) = 2.
  

  log 3500 = log(7×5) + log(100) = log 7 + log 5 + 2 = 0.84510 + 0.69897 + 2

  You can work out the sum! Be sure to notice that it will be more than 3; which makes sense, as the number 3500 is more than 1000.
• One way to work out log 5.5 from the log values we have available, would be to notice that 5.5 = 55 ×10^-1. So its log will be the sum of the logs of those factors:
  

  log(5.5) = log 55 + log(10-1) = log 55 + (-1)

  And we figured out log(55) just a few lines above: it was 0.69897+1.04139. If we subtract 1 from that we will have our answer: 0.69897+0.04139 = 0.74036.
• log 77x² is a product:
  

  log(77x2) = log(7·11·x2) = log 7 + log 11 + 2log x

  You can work out the numerical values. Be sure to notice that log 77 will work out to be more than 1 and less than 2, as 77 is between 10 and 100. And that the 2log x term remains in the answer.

• To evaluate log[(Öx)/6.3] we would notice that Öx can be written as a power of x so that we can use the ''powers'' part of the logarithms rule:
  

  log Öx 6.3 = log x0.5 6.3 = log x0.5 - log 6.3 = (0.5)(log x) - log 6.3

  Next we would find log 6.3. Notice that 6.3 can be factored: 6.3 = 3²·7·10^-1. So its logarithm will be 2log 3 + log 7 + log(10^-1), or 2log 3 + log 7 - 1. Be sure you notice that this number is less than 1: because 6.3 is less than 10.

A thing you should especially notice about this example is that we do not do anything further about the expression ''ln(y+1).'' There are no rules for simplifying logarithms of sums! In particular, ln(y+1) is not the same as ln(y) + ln(1).

• Aw we noted in the problem, the value of A came from weighing the substance at the initial time. That's because q(0) = Ae^0t = Ae⁰ = A·1 = A.
• Now we turn to figuring out the value of k. This is often done by plotting measured values of q(t) versus values of t on special graph paper so that the logarithm of the dependent variable is what gets plotted. The effect is to plot ln q(t) as the dependent variable, where t is the independent variable.

  When we apply the natural log function to q(t) = 400e^kt, we get ln q(t) = ln 400 + ln e^kt = ln 400 + kt.

  When we plot ln q(t) versus t, we get a straight line. Its vertical intercept is ln 400 (which we knew anyway). Its slope is the desired coefficient k.

  Although any two points on a line determine its slope, when we are using lab data we would plot several points to get a reliable value for the line's slope.

• A typical model for exponential decay
  

  q(t) = 650 e-0.07t

  where q(t) is measured in grams and t in hours. We see that this model describes decay rather than growth because of the negative coefficient -0.07 in the exponent.
  1. How much radioactive material was present at the start of the discussion? At the beginning of the discussion, 650 grams of the material present were radioactive. We can see this because when t = 0, q(t) = q(0) = 650 e⁰ = 650.
  2. After 3 hours,
     

     q(3) = 650 e-0.07·3 = 650 e-0.21

     To evaluate q(3) we need a table of exponential values; or (which amounts to the same thing) we could look in a table of natural logarithms, find the number whose logarithm is -0.21, and choose that. Usually we do this by using the table of natural exponential values stored in our calculator: we find that e^-0.21 @ 0.811. Notice that this value is less than 1! ... because the exponent's value is negative.

     So, we find q(3) @ (650)(0.811) @ 526.5

  3. The substance's half-life is the time it takes for its radiactive portion to decrease by half. That would mean that the factor e^-0.07t must be 0.5. In that case:
     

     e-0.07t = 0.5 ln(e-0.07t) = ln(1/2) = ln(1) - ln(2) = -ln(2) -0.07t = -ln(2) t = -ln(2) -0.07 @ 0.693/0.07 @ 9.9

     The half-life of this substance is (approximately) 9.9 hours, which would be (approximately) 9 hours 54 minutes.

• We have put $2500 in a bank account that pays 5 percent interest at the end of each year.
  1. If we put $2500 in the bank at 5 percent interest, at tne end of one year we will have our original $2500 plus interest: (0.05)(2500) = 125. As a percentage, we will have 105 percent of our original amount:
     

     2500 + (0.05)(2500) = (2500)(1 + 0.05) = (2500)(1.05)

  2. As long as the bank is still paying 5 percent interest ... the balance at the end of any year will be 105 percent of the balance at the beginning of the year. Leaving N dollars in the account for 1 year means that the balance will be multiplied by 1.05. It will be 1.05N.
  3. (2500)(1.05¹⁰), because 10 years have gone by; and with the passage of each year the balance is multiplied by 1.05.
  4. To rewrite 2500(1.05^t) in the ''usual'' exponential form
     

     2500(1.05t) = 2500ekt

     we would need to rewrite 1.05 as a power of e:
     

     2500(1.05t) = 2500ekt 1.05t = ekt ln( 1.05t) = ln(ekt) = kt t·ln(1.05) = kt k = ln(1.05) @ 0.0488

     Or, we might simply have remembered that 1.05 = e^ln(1.05). That's what inverse functions do, after all: each undoes the effect of the other!

     Then,
     

     1.05t = (eln(1.05))t = e(ln(1.05))(t) = ekt

     and so (matching the exponents) k = ln(1.05) @ 0.0488.

  5. Money in this account will triple when the coefficient e^0.0488t is 3. For this to happen, ln(0.0488t) = ln(3), or
     

     0.0488t = ln(3) @ 1.099

     Then,
     

     t = ln(3) ln(1.05 @ 1.099 0.0488 @ 22.52

     We will have to wait 22 years, 6 months and a few days for our money to triple.