ELF.01.7 - Solving Exponential ELF.01.7 - Solving Exponential Equations Using Logarithms Equations Using Logarithms

MCB4U - SantowskiMCB4U - Santowski

(A) Strategies for Solving (A) Strategies for Solving Exponential Equations - Guessing Exponential Equations - Guessing

we have explored a variety of equation solving we have explored a variety of equation solving strategies, namely being able to isolate a strategies, namely being able to isolate a variablevariablethis becomes seemingly impossible for this becomes seemingly impossible for exponential equations like 5exponential equations like 5xx = 53 = 53our earlier strategy was to express both sides of our earlier strategy was to express both sides of an equation with a common base, which we an equation with a common base, which we cannot do with the number 53 and the base of 5cannot do with the number 53 and the base of 5Alternatively, we can simply “guess & check” to Alternatively, we can simply “guess & check” to find the right exponent on 5 that gives us 53 find the right exponent on 5 that gives us 53 we know that 5we know that 522 = 25 and 5 = 25 and 533 = 125, so the = 125, so the solution should be somewhere closer to 2 than 3solution should be somewhere closer to 2 than 3

(B) Strategies for Solving (B) Strategies for Solving Exponential Equations - GraphingExponential Equations - GraphingGoing back the example Going back the example of 5of 5xx = 53, we always = 53, we always have the graphing optionhave the graphing option

We simply graph yWe simply graph y11 = 5 = 5xx

and simultaneously graph and simultaneously graph yy22 = 53 and look for an = 53 and look for an

intersection point intersection point (2.46688, 53)(2.46688, 53)

(C) Strategies for Solving (C) Strategies for Solving Exponential Equations - InversesExponential Equations - Inverseshowever, one general strategy that we saw in however, one general strategy that we saw in solving trig equations in Grade 11, was to use an solving trig equations in Grade 11, was to use an "inverse" operation to isolate a variable"inverse" operation to isolate a variableand so now that we know how to "inverse" an and so now that we know how to "inverse" an exponential expression using logarithms, we will exponential expression using logarithms, we will use the same strategy use the same strategy inverse an exponential inverse an exponential using logarithmsusing logarithms

So then if 5So then if 5xx = 53, then log = 53, then log55(53) = x (53) = x but this but this puts us in the same dilemma as before puts us in the same dilemma as before we we don’t know the power on 5 that gives 53don’t know the power on 5 that gives 53

(D) Strategies for Solving (D) Strategies for Solving Exponential Equations - LogarithmsExponential Equations - Logarithms

So we will use the logarithm concept as we apply So we will use the logarithm concept as we apply another logarithm rule another logarithm rule let’s simply take a common let’s simply take a common logarithm of each side of the equation (loglogarithm of each side of the equation (log1010) (since our ) (since our calculators are programmed to work in base 10)calculators are programmed to work in base 10)

Thus, 5Thus, 5xx = 53 now becomes = 53 now becomes

loglog1010(5(5xx) = log) = log1010(53)(53)

loglog1010(5)(5)xx = log = log1010(53)(53)

xlogxlog1010(5) = log(5) = log1010(53) (using log rules)(53) (using log rules)

x = logx = log1010(53) ÷ log(53) ÷ log1010(5)(5)

x = 2.46688 ….. x = 2.46688 …..

(D) Strategies for Solving Exponential (D) Strategies for Solving Exponential Equations – Natural LogarithmsEquations – Natural Logarithms

In solving 5In solving 5xx = 53, we used a common logarithm = 53, we used a common logarithm (log base 10) to solve the equation(log base 10) to solve the equationOne other common logarithm you will see on One other common logarithm you will see on your calculator is the natural logarithm (ln which your calculator is the natural logarithm (ln which uses a special base of numerical value uses a special base of numerical value 2.71828… which is notated by the letter 2.71828… which is notated by the letter ee so so loglogee(x) = ln(x) )(x) = ln(x) )

Thus, ln5Thus, ln5xx = ln53 = ln53And xln5 = ln53And xln5 = ln53And x = ln53 ÷ ln5 = 2.46688 as beforeAnd x = ln53 ÷ ln5 = 2.46688 as before

(E) Examples(E) Examples

Evaluate logEvaluate log3338 = x38 = xAgain, same basic problem Again, same basic problem we are using a we are using a base in which 38 is an awkward number to work base in which 38 is an awkward number to work with (unlike 9,27,81,243,729……)with (unlike 9,27,81,243,729……)So let’s change the expression to an exponential So let’s change the expression to an exponential equation equation 3 3xx = 38 and this puts us back to the = 38 and this puts us back to the point we were at before with 5point we were at before with 5xx = 53!! = 53!!Thus, logThus, log1010(3)(3)xx = log = log1010(38)(38)And xlog3 = log38 (simply dropping the And xlog3 = log38 (simply dropping the implied base 10)implied base 10)So x = log38 ÷ log3 = 3.31107…..So x = log38 ÷ log3 = 3.31107…..

(F) Applications of Exponential (F) Applications of Exponential EquationsEquations

The half-life of radium-226 is 1620 years. After how many years is only 30 The half-life of radium-226 is 1620 years. After how many years is only 30 mg left if the original sample contained 150 mg?mg left if the original sample contained 150 mg?

Recall the formula for half-life is N(t) = NRecall the formula for half-life is N(t) = N00(2)(2)(-t/h)(-t/h) where h refers to the half-life where h refers to the half-life of the substanceof the substance

or N(t) = Nor N(t) = N00(1+r)(1+r)tt where r is the rate of change (or common ratio of -0.5; and where r is the rate of change (or common ratio of -0.5; and t would refer to the number of “conversion periods – or the number of t would refer to the number of “conversion periods – or the number of halving periods)halving periods)

Therefore, 30 = 150(2)Therefore, 30 = 150(2)(-t/1620)(-t/1620) 30/150 = 0.20 = 230/150 = 0.20 = 2(-t/1620) (-t/1620) log(0.2) = (-t/1620) log (2)log(0.2) = (-t/1620) log (2)log (0.2) ÷ log (2) = -t/1620log (0.2) ÷ log (2) = -t/1620-1620 x log(0.2) ÷ log (2) = t-1620 x log(0.2) ÷ log (2) = tThus t = 3761.5 yearsThus t = 3761.5 years

(G) Internet Links(G) Internet Links

College Algebra Tutorial on Exponential EquatioCollege Algebra Tutorial on Exponential Equationsns (NOTE: this lesson uses natural logarithms to (NOTE: this lesson uses natural logarithms to solve exponential equations)solve exponential equations)

Solving Exponential Equations Lesson from PurSolving Exponential Equations Lesson from Purple Mathple Math (NOTE: this lesson uses natural logarithms to (NOTE: this lesson uses natural logarithms to solve exponential equations)solve exponential equations)

SOLVING EXPONENTIAL EQUATIONS from SSOLVING EXPONENTIAL EQUATIONS from SOS MathOS Math (NOTE: this lesson uses natural logarithms to (NOTE: this lesson uses natural logarithms to solve exponential equations)solve exponential equations)

(H) Homework(H) Homework

Nelson text, p132, Q2-12,14,16Nelson text, p132, Q2-12,14,16

AW text, p396, Q12,16,17,18,20,21,24AW text, p396, Q12,16,17,18,20,21,24