1. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
§9.5a Exponential Eqns
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. 9.4 Review § Any QUESTIONS About Any QUESTIONS About HomeWork
MTH 55
Review §
Any QUESTIONS About
§9.4 → Logarithm Change-of-Base
Any QUESTIONS About HomeWork
§9.4 → HW-47

3. Summary of Log Rules
For any positive numbers M, N, and a with a ≠ 1

4. Typical Log-Confusion
Beware that Logs do NOT behave Algebraically. In General:

5. Solving Exponential Equations
Equations with variables in exponents, such as 3x = 5 and 73x = 90 are called EXPONENTIAL EQUATIONS
Certain exponential equations can be solved by using the principle of exponential equality

6. Principle of Exponential Equality
For any real number b, with b ≠ −1, 0, or 1, then
bx = by is equivalent to x = y
That is, Powers of the same base are equal if and only if the exponents are equal

7. Example  Exponential Equality
Solve for x: 5x = 125
SOLUTION
Note that 125 = 53. Thus we can write each side as a power of the same base:
5x = 53
Since the base is the same, 5, the exponents must be equal. Thus, x must be 3. The solution is 3.

8. Example  Exponential Equality
Solve each Exponential Equation
SOLUTION

9. Principle of Logarithmic Equality
For any logarithmic base a, and for x, y > 0,
x = y is equivalent to logax = logay
That is, two expressions are equal if and only if the logarithms of those expressions are equal

10. Example  Logarithmic Equality
Solve for x: 3x+1 = 43
SOLUTION
3 x +1 = 43
Principle of logarithmic equality
log 3 x +1 = log 43
(x +1)log 3 = log 43
Power rule for logs
x +1 = log 43/log 3
x = (log 43/log 3) – 1
The solution is (log 43/log 3) − 1, or approximately

11. Example  Logarithmic Equality
Solve for t: e1.32t = 2000
SOLUTION
e1.32t = 2000
Note that we use the natural logarithm
ln e1.32t = ln 2000
Logarithmic and exponential functions are inverses of each other
1.32t = ln 2000
t = (ln 2000)/1.32

12. To Solve an Equation of the Form at = b for t
Take the logarithm (either natural or common) of both sides.
Use the power rule for exponents so that the variable is no longer written as an exponent.
Divide both sides by the coefficient of the variable to isolate the variable.
If appropriate, use a calculator to find an approximate solution in decimal form.

13. Example  Solve by Taking Logs
Solve each equation and approximate the results to three decimal places.
SOLUTION

14. Example  Solve by Taking Logs
SOLUTION

15. Example  Different Bases
Solve the equation 52x−3 = 3x+1 and approximate the answer to 3 decimals
SOLUTION
Take ln of both sides

16. Example  Eqn Quadratic in Form
Solve for x: 3x − 8∙3−x = 2.
SOLUTION
This equation is quadratic in form.
Let y = 3x then y2 = (3x)2 = 32x. Then,

17. Example  Eqn Quadratic in Form
Soln cont.
But 3x = −2 is not possible because 3x > 0 for all numbers x. So, solve 3x = 4 to find the solution

18. Example  Eqn Quadratic in Form
Soln cont.

19. Example  Population Growth
The following table shows the approximate population and annual growth rate of the United States of America and Pakistan in 2005
Country
Population
Annual Population Growth Rate
USA
295 million
1.0%
Pakistan
162 million
3.1%

20. Example  Population Growth
Use the population model P = P0(1 + r)t and the information in the table, and assume that the growth rate for each country stays the same.
In this model,
P0 is the initial population,
r is the annual growth rate as a decimal
t is the time in years since 2005

21. Example  Population Growth
Use P = P0(1 + r)t and the data table:
to estimate the population of each country in 2015.
If the current growth rate continues, in what year will the population of the United States be 350 million?
If the current growth rate continues, in what year will the population of Pakistan be the same as the population of the United States?

22. Example  Population Growth
SOLUTION: Use model P = P0(1 + r)t
US population in 2005 is P0 = 295. The year 2015 is 10 years from 2005.
Pakistan in 2005 is P0 = 162

23. Example  Population Growth
SOLUTION b.: Solve for t to find when the United States population will be 350.
Some time in yr 2022 ( ) the USA population will be 350 Million

24. Example  Population Growth
SOLUTION c.: Solve for t to find when the population will be the same in both countries.

25. Example  Population Growth
Soln c. cont.
Some time year 2034 ( ) the two populations will be the same.

26. WhiteBoard Work Problems From §9.5 Exercise Set
16, 20, 32, 34, 36, 40
logistic difference equation by Belgian Scientist Pierre Francois Verhulst

27. EMP Widmark BAC Eqn Calculator
All Done for Today
EMP Widmark BAC Eqn Calculator
During the early part of this century, E.M.P. Widmark, a Swedish physician did much of the foundational research regarding alcohol pharmacokinetics in the human body. In addition, he developed an algebraic equation allowing one to estimate any one of six variables given the other five. Typically, we are interested in determining either the amount of alcohol consumed by an individual or the associated blood alcohol concentration (BAC) given the values of the other variables. According to Widmark's equation, the amount of alcohol consumed (A) is a function of these several variables:
Equation 1 N = f(W, r,Ct ,β , t, z)
where: N = amount consumed W = body weight r = the volume of distribution (a constant) Ct = blood alcohol concentration (BAC) β = the alcohol elimination rate t = time since the first drink z = the fluid ounces of alcohol per drink
Widmark's equation relates these variables according to:
Equation 2 N = Wr(Ct + βt) 0.8z
where: N = the number of drinks consumed W = body weight in ounces r = volume of distribution (a constant relating the distribution of water in the body in L/Kg) Ct = the blood alcohol concentration (BAC) in Kg/L β = the alcohol elimination rate in Kg/L/hr t = time since the first drink in hours z = the fluid ounces of alcohol per drink 0.8 = the density of ethanol (0.8 oz. per fluid ounce)
Example 1 Assume that we are interested in determining the amount of alcohol consumed (number of drinks) given certain information. The information we are given includes: a male weighing 185 lbs., r = 0.68 L/Kg, Ct = 0.15 g/100ml, β = g/100ml/hr, t = 5 hours, and drinking 12 fl.oz. beers with 4% alcohol by volume. We introduce this information into Equation 2 according to:
N = (180lb)(16oz/lb)(0.68L/Kg)(0.0015Kg/L + ( Kg/L/hr)(5hr)) (0.8)(0.48fl.oz./drink)
Notice that we had to convert the 0.15 g/100ml and the g/100ml/hr to Kg/L which simply amounts to moving the decimal two places to the left. Solving for A we find:
N = ( ) = 11.5 drinks 0.384
Example 2 The next most common use of Widmark's equation is to determine the blood alcohol concentration (Ct) given the number of drinks consumed. We now assume the following: a female weighing 125 lbs., r = 0.55 L/Kg, β = g/100ml/hr, t = 4 hours, and consumed 7 one fluid ounce glasses of 80 proof vodka. Employing Equation 1 above we introduce the information we are given and solve for Ct as follows:
7 = (125lb)(16oz/lb)(0.55L/Kg)(Ct + ( Kg/L/hr)(4hr)) (0.8)(0.40fl.oz./drink)
Notice that we are interested in finding Ct so we rearrange this equation as follows:
2.24 = 1100(Ct )
Ct = Kg/L = g/100ml
Uncertainty in Widmark Estimates Since each variable introduced into Widmark's equation is subject to measurement uncertainty, the computed values will also have uncertainty propagated through the computations. Widmark developed two other equations (one for men and one for women) where he was able to estimate the uncertainty (one standard deviation) in his computed value for N. His uncertainty equations are (1):
men SN = / N (N (0.68C t W/0.8z))2
women SN = /0.01N (N (0.55C t W/0.8z))2
Applying the equation for men to our Example 1 above we obtain:
SN = / (11.5) ((11.5) (0.68)(0.0015)(180)(16)/(0.8)(0.48) )2
Solving for SN we obtain:
SN = 1.7 drinks
This indicates that the one standard deviation uncertainty for N is 1.7 drinks and the more commonly employed two standard deviation estimate would be 3.4 drinks. Thus, our estimate for the number of drinks in Example 1 above should properly be stated as:
11.5 ± 3.4 drinks
This is a large and overly conservative estimate of the error in N. Other work (2,3) suggests that a better estimate of the two standard deviation estimate in N is closer to ± 20%. For Example 1 the better estimate would be:
11.5 ± 2.3 drinks
Using Breath Alcohol Results in Widmark's Equation Most forensic cases utilizing Widmark's equation will employ breath alcohol rather than blood alcohol results. When doing so, the breath alcohol concentration (BrAC) must be converted to an estimated blood alcohol concentration (BAC) before introducing the value into the equation. This introduces another factor having uncertainty - the BAC/BrAC conversion factor. Typically, the breath and corresponding blood alcohol values are assumed to be the same (using a conversion factor of Kbac/brac = 2100). This is reasonable in forensic cases since it will typically benefit the defendant by providing an underestimate of their true BAC by substituting their BrAC. These principles should be kept in mind when working with Widmark's equation.
References 1. Widmark, E.M.P., Principles and Applications of Medicolegal Alcohol Determination, Davis, CA: Biomedical Publications, 1981, pp
2. Alha, A. R., "Blood Alcohol and Clinical Inebriation in Finnish Men: A Medico-legal Study", Ann. Acad. Scientarum Fennicae, Series A V. Medica-Anthropologica, Vol.26, 1951, pp
3. Gullberg, R.G. and Jones, A.W., "Guidelines for Estimating the Amount of Alcohol Consumed From a Single Measurement of Blood Alcohol Concentration: Re-Evaluation of Widmark's Equation", Forensic Science International, Vol.69, 1994, pp

28. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer –