Lecture: Exponential & Logarithmic Equations & Applications
Lecture script · about 15–20 minutes spoken · Week 15
Opening
A bank customer wants to know how many years a savings balance takes to double. A nursing student wants to know how many hours a dose takes to fall by half. A biology lab wants to know when a culture will reach the count it needs. Each of them knows the start, the target, and the rule. What they don't know is the time, and the time sits up in an exponent. Getting it down is the subject of this lecture.
Here's where we are in this course. In Week 1 you learned the exponent rules and the powers of two, from 1 up to 1024. In Week 2 you learned to check a solution by substituting it back. In Week 3 you met the domain, and in Weeks 11 and 12 you met extraneous solutions, candidates that correct steps produce but the original equation rejects. In Week 13 you built exponential models, and in Week 14 you learned that a logarithm is the exponent you need. This week puts those pieces together so you can solve for an exponent. After this, Week 16 is the finals week, with no new material.
I'll take nine ideas in order, from matching bases to a half-life, and then look back over the whole course.
1. Match the bases
Picture two stacks of identical blocks that stand exactly the same height. They must hold the same number of blocks. Powers of the same base behave that way.
An exponential equation is an equation with the variable in an exponent, such as 2 to the x equals 16. The same-base property says that if two powers with the same base are equal, their exponents are equal.
So take 2 to the x equals 16. From the powers of two, 2 to the fourth is 16. Rewrite the equation as 2 to the x equals 2 to the fourth. Same base, so x is 4. Check: 2 times 2 times 2 times 2 is 16.
Now 3 to the x plus 1 equals 27. Since 3 cubed is 27, the exponents match: x plus 1 equals 3, so x is 2. And 9 to the x equals 3 to the x plus 4 looks harder, because the bases differ. But 9 is 3 squared. By the power rule, 9 to the x becomes 3 to the 2x. Now 2x equals x plus 4, so x is 4. Check: 9 to the fourth is 6561, and 3 to the eighth is 6561 too.
Pause and try this. A classmate solves 2 to the x equals 16 by dividing 16 by 2 and gets 8. Is that right?
Test it before you move on.
Here's the answer. No; 2 to the eighth is 256, not 16, because the x is an exponent, not a factor.
2. Isolate the power before you match
A number in front of a power is like a cover on a box. You take the cover off before you look inside. To isolate the power means to divide both sides by the number multiplying it, so the power stands alone.
Here's a lab culture that starts with 500 cells and doubles every hour. After t hours it holds 500 times 2 to the t cells. When does it reach 4,000 cells? Set 500 times 2 to the t equal to 4,000. Divide both sides by 500, and you get 2 to the t equals 8. That's 2 cubed, so t is 3 hours. Check: 500 times 8 is 4,000.
Money works the same way. In Week 13, 1,000 dollars at 6% compounded once a year grew by a factor of 1.06 each year. When does it reach 1,123.60 dollars? Divide by 1,000 and the right side becomes 1.1236, which is 1.06 squared. So t is 2 years, matching the Week 13 balance table.
Pause and try this. In 500 times 2 to the t equals 4,000, should you multiply 500 by 2 first, or divide by 500 first?
Decide, and then read on.
Here's the answer. Divide by 500 first, because the exponent belongs to the 2 alone, so the power comes before the multiplication.
3. Take a logarithm of both sides
Now the culture needs 5,000 cells. Divide by 500, and you get 2 to the t equals 10. But 10 isn't a power of 2. You can still trap it: 2 cubed is 8 and 2 to the fourth is 16, so t is somewhere between 3 and 4.
To finish, you take a logarithm of both sides: you apply the same logarithm, ln or log, to each side, so the power property can bring the exponent down as a multiplier. Watch it with 2 to the x equals 10. Take ln of both sides. The power property turns the left side into x times ln 2. So x times ln 2 equals ln 10, and dividing gives x equals ln 10 over ln 2. That's the exact answer. A calculator says it's about 3.3219, right between 3 and 4. For the culture, that's about 3.32 hours.
You'll also see that number written as log base 2 of 10. The change-of-base formula says log base b of x equals ln x over ln b, which is how a calculator with only ln and log keys handles base 2.
There's one trap here. Some people take ln on the left and log on the right. That gives x equals 1 over ln 2, about 1.44, and 2 to the 1.44 is about 2.7, nowhere near 10.
Pause and try this. If you take log, base 10, of both sides of 2 to the x equals 10 instead of ln, do you get a different value of x?
Think about what stays the same.
Here's the answer. No; you get x equals 1 over log 2, which is the same 3.3219, because either logarithm works when you use it on both sides.
4. Solve a logarithmic equation by converting it
A logarithm asks a question: what exponent do I need? When an equation hands you that exponent, you can rewrite it as a power. A logarithmic equation is an equation with the variable inside the argument of a logarithm.
Take log base 2 of x minus 1 equals 3. Start with the domain. In Week 14 you learned the argument must be positive, so x minus 1 has to be greater than 0, which means x is greater than 1. Now convert. Log base 2 of x minus 1 equals 3 means 2 cubed equals x minus 1. 2 cubed is 8, so 8 equals x minus 1, and x is 9. That's bigger than 1, so it's allowed. Check: log base 2 of 8 is 3, because 2 cubed is 8.
Pause and try this. Someone sets x minus 1 equal to 3 and gets x equals 4. What went wrong?
Hold that thought for a moment.
Here's the answer. The 3 is the exponent, not the argument, so 2 cubed, which is 8, must equal x minus 1, giving 9.
5. Condense, convert, and run the domain check
Some equations carry two logs. You can't convert two at once, so first you merge them with the product property. Then you convert. Then you check.
Take log of x plus log of x minus 3 equals 1, with log meaning base 10. The domain comes first: both x and x minus 3 must be positive, so x must be greater than 3. Condense: log of x times x minus 3 equals 1. Convert: x times x minus 3 equals 10 to the first, which is 10. Multiply out and set it to 0, and you get x squared minus 3x minus 10 equals 0. That factors as x minus 5 times x plus 2, so the candidates are 5 and negative 2.
Here comes the domain check: the step of testing each candidate in every original logarithm's argument and rejecting any candidate that makes an argument zero or negative. At 5, log of 5 plus log of 2 is log of 10, which is 1. It works. At negative 2, the equation asks for log of negative 2. Every power of 10 is positive, so that log has no real value. Negative 2 is extraneous.
Pause and try this. The quadratic gave two candidates. Why does the original equation keep only one of them?
Give it a moment.
Here's the answer. Negative 2 would put a negative number inside a logarithm, and no power of 10 is negative, so only 5 survives.
6. Find a doubling time
Back to the bank customer. You put 1,000 dollars in an account at 6% compounded continuously. The continuous rate is the constant k in A equals A-naught times e to the kt; here k is 0.06. When is the balance 2,000 dollars?
Set 1,000 times e to the 0.06t equal to 2,000. Divide by 1,000, and e to the 0.06t equals 2. Take ln of both sides, and 0.06t equals ln 2. With ln 2 about 0.6931, t is about 11.55 years. That's the doubling time: the time it takes a growing amount to become twice as large. For continuous growth it's ln 2 over k, whatever the starting amount.
That last phrase matters. Start with 5,000 dollars instead, and the target is 10,000. Divide by 5,000, and you're back at e to the 0.06t equals 2. Same 11.55 years. At 8%, the doubling time drops to about 8.66 years. And the rule of 72, a mental estimate, says 72 over 6, or 12 years, close to the exact 11.55.
Pause and try this. Does 5,000 dollars take longer to double than 1,000 dollars at the same 6%?
Picture the division step before you answer.
Here's the answer. No; the starting amount divides out, so both double in about 11.55 years.
7. Find a tripling time
Doubling is the famous question, but tripling works the same way. The tripling time is the time it takes a growing amount to become three times as large, ln 3 over k for continuous growth.
Take 2,000 dollars at 5% compounded continuously. When is it 6,000 dollars? Divide by 2,000, and e to the 0.05t equals 3. Take ln, and 0.05t equals ln 3. With ln 3 about 1.0986, t is about 21.97 years. The 2,000 divided out, so any amount at 5% triples in that same time.
Pause and try this. Which number replaced the 2 from the doubling-time steps when the goal became tripling?
Run through the steps in your head.
Here's the answer. A 3 replaced the 2, so ln 3 sits where ln 2 sat, and the time is ln 3 over 0.05.
8. Find a half-life
Decay runs the same machine downhill. A lab sample loses mass continuously, following A equals A-naught times e to the negative 0.15t, with t in hours. So k is negative 0.15 per hour, and the minus sign is what makes it decay. In Week 13 you defined a half-life as the time it takes a decaying amount to fall to half its size.
Set the amount equal to half of A-naught. Divide by A-naught, and e to the negative 0.15t equals one half. Take ln of both sides. The ln of one half is negative ln 2, so negative 0.15t equals negative ln 2. The two minus signs cancel, and t is ln 2 over 0.15, about 4.62 hours.
Now a 400-gram sample. How long until 50 grams are left? Halve it: 400, 200, 100, 50. That's three half-lives, so 3 times 4.62, which is 13.86 hours.
Pause and try this. If someone drops the minus sign and writes e to the 0.15t, what goes wrong?
Think about which way the amount moves.
Here's the answer. The model would grow instead of decay, because a positive k means growth, so the minus sign has to stay attached to 0.15.
9. Look back: one habit, fifteen weeks
Every solve in this lecture used something from earlier in the course. The Week 1 power rule rewrote 9 as 3 squared. The Week 2 check by substitution caught the divide-by-2 trap. The Week 3 domain became the domain check. Week 6 factoring and Week 7 quadratics solved x squared minus 3x minus 10 equals 0. Weeks 11 and 12 taught you to distrust a candidate until it passes the original equation.
So this course was never a pile of separate tricks. It was a few moves you can always check: undo an operation, keep the equation balanced, and substitute to confirm.
Pause and try this. In the log equation with candidates 5 and negative 2, which earlier week's habit rejected negative 2?
Name the week before you read on.
Here's the answer. The domain idea from Week 3, joined to the extraneous-solution habit from Weeks 11 and 12, rejected negative 2, because it puts a negative number inside a log.
Closing
Let's pull it together, in the order we took it. When both sides are powers of the same base, set the exponents equal. Divide away a coefficient before you do anything else. When the bases won't match, take the same log of both sides, so 2 to the x equals 10 gives ln 10 over ln 2. Convert a log equation to a power, and condense first if there are two logs. Write the domain, and reject any candidate that puts a zero or a negative inside a log. Doubling time is ln 2 over k, tripling time is ln 3 over k, and half-life is ln 2 over 0.15 for our sample. The starting amount never changes them.
Next in this module, you'll find the slide deck, then the Readings & Resources page, the lecture tutorial, the practice set, and the quiz. After that comes Week 16, the finals week, where every move from this course comes back together.
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