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0.9 Logarithms  (Page 4/4)

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Interest calculations can be quantified by mathematical formulas. Suppose that we are concerned with making an investment of P dollars at an annual rate of interest of i for n years. Here P denotes the present value of the investment, n represents the number of years the money is to be invested, and i is the interest rate per annum. Let us denote by F the future value of the investment at the end of n years. The value of F can be calculated via the formula

F = P ( 1 + i ) n size 12{F=P` \( 1+i \) rSup { size 8{n} } } {}

Let us illustrate the use of this formula by means of a problem.

Question: A couple has just had a baby. The couple wishes to make an investment in an account that will be used to fund college expenses. The couple visits an investment advisor and establishes a tax-free college savings account. By tax-free, we mean an account where income tax is not paid yearly. The couple is advised that this feature will enable the account to grow faster.

The couple deposits $25,000 in the account. The couple is told that the value of the account will appreciate at an annual rate of 8% . What will be the value of the account when the child turns 18 years of age?

Solution: We are asked to find the value of F . From the problem statement, P is $25,000, i is 0.08 and n is 18. We substitute into the interest formula to obtain the result

F = ( 25 , 000 ) ( 1 + 0 . 08 ) 18 = ( 25 , 000 ) ( 1 . 08 ) 18 = ( 25 , 000 ) ( 3 . 996 ) = 99 , 900 size 12{F= \( "25","000" \) ` \( 1+0 "." "08" \) rSup { size 8{"18"} } = \( "25","000" \) ` \( 1 "." "08" \) rSup { size 8{"18"} } = \( "25","000" \) ` \( 3 "." "996" \) ="99","900"} {}

We conclude that the account will be worth $99,900 at the end of 18 years.

Let us now a related problem that involves the use of logarithms.

Question: An individual inherits $10,000 from a relative. The individual wishes to invest the sum in a tax-free account. He/she plans to use the proceeds of this account as a down payment on a future purchase of a home. The annual interest rate of the account is 6%,

The individual anticipates that he/she will need at least $20,000 for the down payment. How long will it take for the value of the account to grow to $20,000?

Solution: Once again, we begin with the identification of the parameters of the problem. Here, P is 10,000, i is 0.06, and F is 20,000. We incorporate these values into the interest formula

20 , 000 = ( 10 , 000 ) ( 1 + 0 . 06 ) n size 12{"20","000"= \( "10","000" \) ` \( 1+0 "." "06" \) rSup { size 8{n} } } {}

Let us divide each side by (10,000) and re-arrange terms

( 1 . 06 ) n = 2 size 12{ \( 1 "." "06" \) rSup { size 8{n} } =2} {}

Now let us take the logarithm of each side of the equation. We will use 1.06 as the base of the logarithm

n = log 1 . 06 ( 2 ) size 12{n="log" rSub { size 8{1 "." "06"} } \( 2 \) } {}

The base 1.06 logarithm is related to the base 10 logarithm as follows

log 1 . 06 ( x ) = log 10 ( x ) log 10 ( 1 . 06 ) size 12{"log" rSub { size 8{1 "." "06"} } \( x \) = { {"log" rSub { size 8{"10"} } \( x \) } over {"log" rSub { size 8{"10"} } \( 1 "." "06" \) } } } {}

We will use this relationship to help us find the value for n

n = log 10 ( 2 ) log 10 ( 1 . 06 ) = 0 . 301 0 . 0253 = 11 . 90 size 12{n= { {"log" rSub { size 8{"10"} } \( 2 \) } over {"log" rSub { size 8{"10"} } \( 1 "." "06" \) } } = { {0 "." "301"} over {0 "." "0253"} } ="11" "." "90"} {}

So the individual should plan on waiting 11.90 years for the account to grow to a value of $20,000.

Problems such as the previous one often make use of the relationship

log a ( x ) = log b ( x ) log b ( a ) size 12{"log" rSub { size 8{a} } \( x \) = { {"log" rSub { size 8{b} } \( x \) } over {"log" rSub { size 8{b} } \( a \) } } } {}

This relationship allows one to convert from the logarithm of any base to a logarithm of another base. Typically, scientific calculators are only able to compute base 10 and natural (base e ) logarithms. One should become acquainted with the conversion of logarithms of other bases to base 10 or base e logarithms.

Exercises

  1. The power of the signal entering an amplifier is 15 mW. The power of the signal that leaves the amplifier is 25 W. Express the gain of the amplifier in decibels.
  2. Unlike an amplifier, some devices reduce the power of input signals. This process is called attenuation. Suppose that the power that enters an attenuator is 60 W and the power at the output is 0.9 W. Express the gain of the attenuator in decibels.
  3. Consider a signal processing scheme such as that shown in Figure 2. The power of the input signal before it passes through the noisy channel is 40 W. After passing through the noisy channel, the original signal is corrupted by noise. The noise component has a power of 10 W. What is the Signal to Noise ratio of the signal that emerges from the noisy channel?
  4. Consider the situation described in exercise 3. The noisy signal enters a signal processor. The signal processor diminishes the noise power of the signal by 5%, while diminishing the power in the noise component by 95%, What is the SNR of the output of the signal processor?
  5. Consider the circuit shown in Figure 3. Let us replace the resistor with another whose value is 200 kΩ. (a) What is the new value of the time constant of the circuit. (b) Find the value of the transient response when t = 0.8 seconds. (c) Determine the value of time at which the transient response decays to a value of 1 Volt.

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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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