Rational Exponents and More Word Problems

Exponent Laws Review When multiplying powers with the same base, we keep the base and add the exponents. The value of any power with exponent 0 is 1. When dividing powers with the same base, we keep the base and subtract the exponents. To change the sign of an exponent, change the base to its reciprocal.(To change the base to its reciprocal, change the sign of the exponent) When we have a power of a power, we keep the base and multiply the exponents.

Rational exponents A rational number is one that can be expressed as a fraction The numbers π and are examples of irrational numbers. Ex. Solve for x. We know to solve this we must “undo” the square by taking the square root of both sides. 1 But what exactly is the square root? In order for the square root to undo the exponent 2, it must also be an exponent. Notice that in the final answer the x term has an exponent of 1. So the problem is actually solved like this…

Rational exponents Ex. Solve for x. We raise both sides to some exponent, ? But what is this exponent? We see that one the left-hand side we have a power of a power so we will multiply the exponents to get 1. (2)(?) = 1 The missing exponent must be 1/2 ! So the square root is actually an exponent of 1/2

Rational exponents So what does an exponent of 1/3 mean? This means that an exponent of 1/3 is the same as the cubed-root. It is like asking what cubed gives me this base? Ex. Solve for x: 81/3 = x We could cube both sides. The left-hand side is a power of a power so we multiply the exponents. We know that 23 is 8 so x = 2. In fact all the rational exponents are like this. Ex. Evaluate: a) b) c) a) 2 (26 = 64) b) 3 (34 = 81) c. 5 (53 = 125)

Rational exponents Remember that this all started by recognizing that the square root was an exponent. It follows, then, that all rational exponents can be represented by root signs. Ex. Let’s step it up a notch… in all of these examples the bases are also powers… We can write them as so:

Rational exponents In each case we now have a power of a power, so we can multiply the exponents. So now we can have rational exponents where the numerator is not 1…

Rational exponents Evaluate the following without a calculator. Show one intermediate step. When we need to evaluate an expression with a rational exponent, we can “rip the exponent apart”. The exponent can be viewed as a 2 and a 1/3 However, we are not responsible to know 1252. Is there another way? Since the new expression is a power of a power, we multiply the exponents. It does not matter what order there are in. So let’s switch them. We know that 1251/3 is 5!

Rational exponents Evaluate the following without a calculator. Show one intermediate step. When we need to evaluate an expression with a rational exponent, we can “rip the exponent apart”. The exponent can be viewed as a 3 and a ½ (hmm…) Or The exponent can be viewed as a ½ and a 3. (hmm.. this is very helpful, since we know the square roots of 25 and 4!)

Rational exponents Try these: Evaluate each without a calculator by showing an intermediate step. a) 45/2 b) 81−3/4 c) 10−2.5 d) 82/3 a) b) c) d)

Rational exponents e) f) g) e) f) g)

Solving Exponential Equations (Rational exponents) Remember: An exponential equation is one where the unknown (x) is in the exponent. Ex. Solve: Notice that we do not know the cubed-root of 32 or the square root of 8… but we can get common bases Now consolidate each exponent, remembering that a square root is an exponent of ½ and the cubed-root is the exponent 1/3 Since the bases are now equal, so are the exponents…then cross-multiply and solve

Solving Exponential Equations (Rational exponents) Try this one. Solve for x.

Applications (word problems) We have seen that patterns with a common ratio can be described with an exponential equation. Ex. 120, 60, 30, 15, 7.5… We have also seen that for most applications involving time, we use a more general version: Where: x measures time since the start y is the amount at time x, r is the common ratio, A0 is the original amount (ie, the amount at time x = 0), and period is the amount by which the x values increase. (Note, sometimes x is replaced by t to emphasize that it measures time.) Using the exponential pattern formula we get

Applications (word problems) Ex 1. A certain population has been seen to triple every 12 years. In 1950, there was 2500 individuals. a) What is the population in 2011? We know: A0 = 2500 (initial population…) r = 3 (…which triples…) period = 12 (…every 12 years) We want to find the population (y) in 2011 (when x = 2011 – 1950 = 61years) In 2011, the population was 665 718

Applications (word problems) Ex 1. A certain population has been seen to triple every 12 years. In 1950, there was 2500 individuals. b) When was the population 67 500? We know: A0 = 2500 (initial population…) r = 3 (…which triples…) period = 12 (…every 12 years) We want to find x (number of years since 1950) when y = 67 500 Isolate the power!! Common bases anyone? In 36 years, that is in 1986, the population was 67 500

Applications (% increase) Ex 2. Bob’s salary increases by 6% every 15 years. In 2000, his salary was $35 000. What will it be in 2020? We could start with the equation, but sometimes starting with a table is easier: x (years since 2000) 15 30 45 y (salary, in thousands of dollars) 35 37.1 39.326 41.686 To calculate the salary after 15 years, we can take 6% of the present salary and add it to the present salary… Or Take 106% of the present salary 6% of 35 = 0.06(35) = 2.1 106% of 37.1 = 1.06(37.1) = 39.326 2.1 + 35 = 37.1 106% of 39.33 = 1.06(39.326) = 41.686 106% of 35 = 1.06(35) = 37.1

Applications (% increase) Ex 2. Bob’s salary increases by 6% every 15 years. In 2000, his salary was $35 000. What will it be in 2020? x (years since 2000) 15 30 45 y (salary, in thousands of dollars) 35 37.1 39.326 41.686 ×1.06 ×1.06 ×1.06 Let’s find the equation of this pattern. We’ll need the common ratio. Each term is 1.06 times as big as the last, CR = 1.06 … but of course!

Applications (% increase) Ex 2. Bob’s salary increases by 6% every 15 years. In 2000, his salary was $35 000. What will it be in 2020? x (years since 2000) 15 30 45 y (salary, in thousands of dollars) 35 37.1 39.326 41.686 So the equation for this data is: Plug in 20 for x (notice we don’t plug in 2020) to solve for y. In 2020 Bob’s salary will be $37 830

Shortcut for “% increase” (appreciation) and “% decrease” (depreciation) questions When a value increases or appreciates by a certain percentage, the common ratio will be: r = 1 + (percent increase in decimal form) When a value decreases or depreciates by a certain rate, the common ratio will be: r = 1 − (percent decrease in decimal form) Notice that even when something depreciates in value, r is still positive. if r > 1, the values of y will be increasing if 0 < r < 1, the values of y will be decreasing if r = 1, the values of y will not change (if r < 0, the values of y will alternate between + and −)

Applications (% decrease) Ex 3. In a certain rural town, the population is decreasing by 25% every 6 years. If there are 2300 people in the town this year, how long will it take before there are only 5 people left? Since this is a % decrease question, we can calculate r as follows: r = 1 – 0.25 = 0.75 This means that every 6 years, only 75% of the previous population remains. period = 6 yrs A0 = 2300 y = 5 x = ? We can’t write these with common bases, so we can’t solve this yet. Stay tuned for logarithms which will help us determine that the answer is 127.9 years

Applications (half-life) Another exponential situation involves the half-life of radioactive materials. Certain chemicals are unstable, and decay at a predictable rate. We model this with HALF-LIFE The half life of a substance is the time it takes for half of the original material to decay. After two half-lives for example, only ¼ of the original amount remains. All half-life questions have: r = ½, and period = half life

Applications (half-life) Ex 4. A certain radioactive substance has a half of 8 days. It initially contained 90 mg. When will there be 5.625 mg left? period = half-life = 8 days r = 1/2 A0 = 90 mg y = 5.625 mg x = ? In 32 days, there will be 5.625 mg left.

Applications (half-life) Ex 5. A certain substance has a half-life of 65 minutes. When will there be of the original amount? period = half-life = 8 days r = 1/2 A0 = ? x = ? y = um Notice that we were not given A0, the original amount, but y value is 1/64th of A0, we use y = Divide both sides by A0 After 6 and a half hours, there will remain but 1/64th of the original amount.

Applications (compound interest) Another application of exponential growth is compound interest. Compound interest gives you a set percentage of the amount in your account each compounding period. After the first period, this amount includes some interest already earned. With compound interest you’re getting interest on the interest. This is different than simple interest which simply gives you a set amount period. Compounding periods can be: daily (1/365 of a year) weekly (1/52 of a year) quarterly (1/4 of a year) semi-annually (1/2 of a year) monthly (1/12 of a year) etc. Careful! In these types of questions: x measures the number of years, period is the fraction of a year per compounding period

Applications (compound interest) Ex 6. You invest $100 in an account paying 6% annual interest, compounded quarterly. How much will you have in 10 years? You get 6% (that is 0.06) in the year but you get it spread over 4 compounding periods (quarterly). Each compounding period you’re getting: 6% ÷ 4 = 1.5%, (that is 0.015) Recall that this is a 1.5% increase, so r = 1 + 0.015 = 1.015 period = 1/4 r = 1 + 0.015 = 1.015 A0 = 100 x = 10 y = ?

Applications (compound interest) Ex 6. You invest $100 in an account paying 6% annual interest, compounded quarterly. How much will you have in 10 years? So the equation would look like this: Now we can plug in x =10

Applications (compound interest) Here is a shortcut formula for compound interest: Where A is the $ amount in the account at time t (years) P is the principle (initial) $ amount (when t = 0) i is the decimal value of the annual interest rate n is how many times per year the interest is compounded t is the number of years Look for terms like: daily (n = 365), weekly (n = 52) quarterly (n = 4) semi-annually (n = 2) monthly (n = 12)

Applications (compound interest) Ex 7. An bank account earns interest compounded monthly. The investment doubles in 9.27 years. Calculate the annual interest rate. When the money doubles there will be 2P in the account. So A = 2P We are now solving for the base. We must “undo” the exponent. Raise both sides to the

Applications (compound interest) Ex 8. Which is better: 5% interest per year compounded monthly, or 5% per year compounded daily? Let’s assume an initial investment of $100 and a term of 10 years. Because the interest is compounded more often (even though each time it is a smaller percentage) the account paying the daily compounded interest is better.