RATIONAL EXPONENTS AND MORE WORD PROBLEMS REVIEW Last class we saw that (2x5)(3x3) = 6x8 вЂў Powers with a common base can be easily multiplied and divided вЂў Powers themselves can be raised to another power вЂў Powers with a negative exponent are equivalent to the reciprocal base and positive exponent. вЂў Exponential equations can be solved if the powers contain the same base. 28 / 25 = 23 = 8 (4d6)2 = 16d12 3e-3 = 3 e3 42x = 83 (22)2x = (23)3 24x = 2 9 4x = 9 x = 9/4 RATIONAL EXPONENTS A rational number is one that can be expressed as a fraction a . The numbers ПЂ and в€љ2 are b examples of irrational numbers. Ex. Solve for x. x пЂЅ 25 2 We know to solve this we must вЂњundoвЂќ the square by taking the square root of both sides. 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 вЂ�xвЂ™ term has an exponent of 1. x 2 пЂЅ 25 x 1 пЂЅ п‚±5 So the problem is actually solved like thisвЂ¦ RATIONAL EXPONENTS Ex. Solve for x. x пЂЅ 25 2 пЂЁx пЂ© 2 ? пЂЅ 25 We raise both sides to an 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 is 1/2 !! 1 x пЂЅ п‚±5 пЂЁx пЂ© 2 1 1 2 пЂЅ 25 2 x пЂЅ п‚±5 So the square root is actually an exponent of 1/2 RATIONAL EXPONENTS This means that an exponent of 1/3 is the same as the cubedroot. It is like asking what cubed gives me this base? So what does an exponent of 1/3 mean? Ex. Solve for x 81/3 = x пѓ¦ пѓ§83 пѓ§ пѓЁ 1 3 пѓ¶ пѓ· пЂЅ x3 пѓ· пѓё 8пЂЅ x 3 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. a) 641/6 b) 811/4 c. 1251/3 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. 1 1 5 49 пЂЅ 49 3 5 1 64 пЂЅ 64 3 4 100 пЂЅ 100 In all of these examples the bases are perfect squares. We can write them as so. 5 7 2 пЂЁ пЂ© пЂЅ 7 2 5 пЂЁ пЂ© 1 1 3 8 пЂЅ 8 2 2 3 4 10 2 пЂЁ пЂЅ 10 2 пЂ© 1 4 4 RATIONAL EXPONENTS 5 7 2 пЂЁ пЂ© пЂЅ 7 2 пЂЁ пЂ© 1 1 3 5 8 пЂЅ 8 2 2 4 3 10 2 пЂЁ пЂЅ 10 2 пЂ© 1 4 In each case we have a power of a power and can multiply the exponents. 5 7 2 пЂЁ пЂ© пЂЅ 7 2 2 1 5 пЂЅ7 5 4 3 8 2 пЂЁ пЂ© пЂЅ 8 2 1 2 3 пЂЅ 83 So now we can have rational exponents where the numerator is not 1. 10 2 пЂЁ пЂЅ 10 2 пЂ© 1 1 4 пЂЅ 10 2 RATIONAL EXPONENTS Evaluate the following with a calculator. Show one intermediate step. 2 125 пЂЁ125 пЂ© 2 3 1 3 When we need to evaluate an expression with a rational exponent, we can вЂњrip the exponent apartвЂќ. The exponent can be viewed as a square and an exponent of 1/3 However, we are not responsible to know 1252. Is there another way? пѓ¦ пѓ§ 125 пѓ§ пѓЁ 1 3 пѓ¶ пѓ· пѓ· пѓё 2 We know that 1251/3 is 5! пѓ¦ пѓ§ 125 пѓ§ пѓЁ 1 3 2 пѓ¶ пѓ· пЂЅ 52 пѓ· пѓё Since the new expression is a power of a power, we multiply the exponents. It does not matter what order there are in. So letвЂ™s switch them. 2 125 3 пЂЅ 25 RATIONAL EXPONENTS Evaluate the following with a calculator. Show one intermediate step. 3 пѓ¦ 9 пѓ¶2 пѓ§ пѓ· пѓЁ4пѓё 1 пѓ¦ пѓ¦ 9 пѓ¶3 пѓ¶ 2 пѓ§пѓ§ пѓ· пѓ· пѓ§пѓЁ 4 пѓё пѓ· пѓЁ пѓё пѓ¦ пѓ¶ 2 9 пѓ§пѓ¦ пѓ¶ пѓ· пѓ§пѓ§ 4 пѓ· пѓ· пѓ§пѓЁ пѓё пѓ· пѓЁ пѓё 1 3 When we need to evaluate an expression with a rational exponent, we can вЂњrip the exponent apartвЂќ. The exponent can be viewed as a cube and an exponent of 1/2 The exponent can be viewed as an exponent of ВЅ and a cube ORвЂ¦ And we know the square root of 9 and 4. 3 27 пѓ¦3пѓ¶ пѓ§ пѓ· пЂЅ 8 пѓЁ2пѓё RATIONAL EXPONENTS Try these: 1. Evaluate each without a calculator by showing an intermediate step. a) 45/2 b) 81-3/4 a) пѓ¦ 1 пѓ§42 пѓ§ пѓЁ c) 100-2.5 5 пѓ¶ пѓ· пЂЅ 2 5 пЂЅ 32 пѓ· пѓё c) пЂ5 100 2 пѓ¦ пЂЅ пѓ§ 100 пѓ§ пѓЁ 1 2 пѓ¶ пѓ· пѓ· пѓё b) пѓ¦ 1 пѓ§ 81 4 пѓ§ пѓЁ пѓ¶ пѓ· пѓ· пѓё пЂ5 пЂЅ пЂ3 пЂЅ3 пЂ3 пЂЅ 1 27 пЂ5 пЂЅ 10 1 10 5 пЂЅ 1 100000 RATIONAL EXPONENTS Try these: 1. Evaluate each without a calculator by showing an intermediate step. d) 3 d) 125 3 пЂЁ 3 пѓ¦ e) пѓ§ пѓ§ пѓЁ пЂ1 125 пЂ1 пЂ© пЂ1 125 пЂЁ5 пЂ© e) пѓ¦ пѓ§ пѓ§ пѓЁ 4 пѓ¶ пѓ· 49 пѓ·пѓё 4 пѓ¶ пѓ· 49 пѓ·пѓё пЂ1 пЂ1 пЂ1 f) пЂЁ 8 f) пЂЁ 8 3 3 пЂ© пЂ5 пЂ© пЂЁ 8пЂ© пЂ5 3 1 пѓ¦2пѓ¶ пѓ§ пѓ· пѓЁ7пѓё 5 7 2 2 1 пЂ1 пЂ5 2 пЂ5 1 5 32 RATIONAL EXPONENTS Try these: 2. Solve the exponential equation пЂЁ 8пЂ© xпЂ«4 пЂЅ Notice that we do not know the cubed-root of 32 or the square root of 8. 1 3 32 But we can get COMMON BASES пЂЁ2пЂ© xпЂ«4 3 пѓ¦ 3 пѓ§ 2 пѓЁ пЂЁ пЂ© пѓ¦ пѓ§22 пѓ§ пѓЁ 3 1 2 пЂЅ 3 1 2 пѓ¶ пѓ· пѓё пѓ¶ пѓ· пѓ· пѓё xпЂ«4 xпЂ«4 Now simplify each side, remembering that a square root is an exponent of ВЅ and the cubed-root is the exponent 1/3 5 1 Power of a powerвЂ¦. пѓ¦ 1 пѓ¶3 пЂЅпѓ§ 5пѓ· пѓЁ2 пѓё пЂЁ пЂ© пЂЅ 2 пЂ5 1 3 пѓ¦ пѓ§2 пѓ§ пѓЁ 3 x пЂ« 12 2 пЂ5 пѓ¶ пѓ· пЂЅ пЂЁ2 пЂ© 3 пѓ· пѓё Since the bases are equal so are the 3 x пЂ« 12 exponents. 2 пЂЅ пЂ5 3 RATIONAL EXPONENTS 3 x пЂ« 12 пЂЅ пЂ5 2 We can (finally) solve for x. 3 3 x пЂЅ пЂ 3 . 33 пЂ 12 пѓ¦пЂ5пѓ¶ 3 x пЂ« 12 пЂЅ 2 пѓ§ пѓ· пѓЁ 3 пѓё 3x пЂЅ пЂ 10 3 x пЂЅ пЂ 15 . 33 x пЂЅ пЂ 5 . 11 пЂ 12 3 2 4пЂ x 1 3. Solve for x. 9 3 пЂЅ 27 пЂЁ3 пЂ© 2 4пЂ x 3 9 пЂЅ 27 1 3 33 пЂЅ 3 2 3 пЂЅ пЂ© 4пЂ x 6 12 пЂ 3 x 6 6 2*6 12 пЂ 3 x 2 6 3 12 пЂ 3 x 3 6 пЂЅ пЂЁ3 пЂЅ 6 пЂЅ 12 пЂ 3 x 3 4 пЂЅ 12 пЂ 3 x пЂ 8 пЂЅ пЂ3x xпЂЅ 8 3 BACK TO SOME WORD PROBLEMS We have seen that patterns with a common ratio can be described with an exponential equation. Ex. 120,60,30,15,7.5вЂ¦ This can be written as t n пЂЅ 120 ( 0 . 5 ) n пЂ1 We have also seen that when a data table is given we can use this general equation: x y пЂЅ a (r ) period Where вЂ�aвЂ™ is the initial amount (at the beginning) and period is when the value of y increases by r times MORE WORD PROBLEMS x y пЂЅ a (r ) period Where вЂ�aвЂ™ is the initial amount (at the beginning) and period is when the value of y increases by r times Ex 1. A certain population has been seen to triple every 12 years. In 1950, there was 2500 individuals. How many was there in 2011? The equation is: x y пЂЅ 2500 ( 3 ) 12 61 y пЂЅ 2500 ( 3 ) 12 y пЂЅ 2500 ( 3 ) We need to solve for y knowing that x = 2011-1950 5 . 0833 y пЂЅ 2500 ( 266 . 2873 ) y пЂЅ 665718 . 28 In 2011 there was 665718 individuals. MORE WORD PROBLEMS x y пЂЅ a (r ) period Where вЂ�aвЂ™ is the initial amount (at the beginning) and period is when the value of y increases by r times Ex 2. A certain population has been seen to triple every 12 years. In 1950, there was 2500 individuals. When will there be 67500 individuals? The equation is: We need to solve for x knowing that y = 67500 x y пЂЅ 2500 ( 3 ) 12 Isolate the power!! Common bases anyone? x 67500 пЂЅ 2500 ( 3 ) 12 27 пЂЅ ( 3 ) x x 12 3 пЂЅ ( 3 ) 12 3 3пЂЅ In 36 years (1986) there would be 67500 individuals. x 12 x пЂЅ 36 MORE WORD PROBLEMS Ex 3. A teacherвЂ™s salary increases by 6% every 15 years. In 2000, a starting teacherвЂ™s salary was $35000. What will it be in 2020? We can start this by looking at a table of values. x (years since 2000) 0 Starting salary (in thousands of 35 dollars) 15 37.1 30 45 39.33 41.69 IвЂ™m rich! To calculate the salary 6% of 35 after 15 years, we can 2.1 + 35 = 37.1 = 0.06(35) take 6% of the present = 2.1 salary and add it to the present salary. Now take 6% of 37.1 and add it to 37.1 Now take 6% of 39.33 and add it to 39.33 MORE WORD PROBLEMS Ex 3. A teacherвЂ™s salary increases by 6% every 15 years. In 2000, a starting teacherвЂ™s salary was $35000. What will it be in 2020? We can start this by looking at a table of values. x (years since 2000) 0 15 Starting salary (in thousands of 35 dollars) 37.1 1.06 Now we can find the equation. So letвЂ™s find the common ratio. 30 45 39.33 41.69 1.06 1.06 6% of x = 0.06x 0.06x + x = 1.06x Each term is 1.06 times as big as the last Of course! We got each successive value by taking 6% of the last value and adding it to the last value. MORE WORD PROBLEMS Ex 3. A teacherвЂ™s salary increases by 6% every 15 years. In 2000, a starting teacherвЂ™s salary was $35000. What will it be in 2020? We can start this by looking at a table of values. x (years since 2000) 0 Starting salary (in thousands of 35 dollars) When a value increases or appreciates by a certain rate, the common ratio will be 1 + rate/100 15 37.1 30 45 39.33 41.69 When a value decreases or depreciates by a certain rate, the common ratio will be 1 - rate/100 MORE WORD PROBLEMS Ex 3. A teacherвЂ™s salary increases by 6% every 15 years. In 2000, a starting teacherвЂ™s salary was $35000. What will it be in 2020? We can start this by looking at a table of values. x (years since 2000) 0 15 Starting salary (in thousands of 35 dollars) 37.1 30 45 39.33 41.69 So the equation for this data is x 20 y пЂЅ 35 (1 . 06 ) 15 y пЂЅ 35 (1 . 06 ) 15 Plug in 20 for x (years since 2000) to solve for y y пЂЅ 37 . 83 In 2020 a starting teacher will make $37 830 LI OR FE Another exponential situation involves radioactive material. Certain isotopes of the elements that make up the world around us are unstable. Each nucleus decays randomly but overall there is a pattern. The half life of a We model this with HALF-LIFE substance is the time for half of the material to decay into something else. Ex. A certain radioactive After two half-lives, Вј of substance has a half of 8 the original amount days. It initially contained remains. 90 mg. When will there be 5.625 mg left? All half-life questions have a common ratio (and base) of ВЅ. In 32 days, there will be 5.625 mg left. x x пѓ¦ 1 пѓ¶8 y пЂЅ 90 пѓ§ пѓ· пѓЁ2пѓё пѓ¦ 1 пѓ¶8 5 . 625 пЂЅ 90 пѓ§ пѓ· пѓЁ2пѓё x пѓ¦ 1 пѓ¶8 пЂЅпѓ§ пѓ· 16 пѓЁ 2 пѓё 1 4 x пѓ¦1пѓ¶ пѓ¦ 1 пѓ¶8 пѓ§ пѓ· пЂЅпѓ§ пѓ· пѓЁ2пѓё пѓЁ2пѓё 4пЂЅ x 8 x пЂЅ 32 LI OR FE Ex 2. A certain substance has a half-life of 65 minutes. When will there be 1 th of the original amount? 64 However, we can still solve the problem. The equation would be x Notice that we were not given the original amount. пѓ¦ 1 пѓ¶ 65 y пЂЅ aпѓ§ пѓ· пѓЁ2пѓё x пѓ¦ 1 пѓ¶ 65 a пЂЅ aпѓ§ пѓ· 64 пѓЁ2пѓё 1 The y value is 1/64 times вЂ�aвЂ™ or 1 ( a ) 64 x пѓ¦1пѓ¶ пЂЅпѓ§ пѓ· 64 пѓЁ 2 пѓё 1 6 65 The next step would be to divide both sides by вЂ�aвЂ™. x пѓ¦1пѓ¶ пѓ¦ 1 пѓ¶ 65 пѓ§ пѓ· пЂЅпѓ§ пѓ· пѓЁ2пѓё пѓЁ2пѓё 6пЂЅ x 65 x пЂЅ 390 INTEREST INTEREST Another type of exponential equation is the compound interest equation. This is different than simple interest which simply gives you a set amount each time. With compound interest youвЂ™re getting interest on the interest. Compound interest gives you a set percentage of the amount in your account. This amount may include some interest already earned. Ex. You invest $100 in an account paying 6% interest a year compounded quarterly. How much will you have in 10 years? You get 6% in the year but you get it spread over 4 times (quarterly). Each time you get an interest payment youвЂ™re getting 6%/4 = 1.5% INTEREST INTEREST Ex. You invest $100 in an account paying 6% interest a year compounded quarterly. How much will you have in 10 years? So the equation would look like this: Simplifying the exponent gives 0 . 06 пѓ¶ пѓ¦ y пЂЅ 100 пѓ§ 1 пЂ« пѓ· 4 пѓё пѓЁ Now we can plug in x =10 x 0 . 06 пѓ¶ 14 пѓ¦ y пЂЅ 100 пѓ§ 1 пЂ« пѓ· 4 пѓё пѓЁ 4x 0 . 06 пѓ¶ пѓ¦ y пЂЅ 100 пѓ§ 1 пЂ« пѓ· 4 пѓё пѓЁ The amount of interest earned each period 4 (10 ) y = $181.40 x is time in years. The period is Вј of a year INTEREST INTEREST In general, the compound interest is rпѓ¶ пѓ¦ A пЂЅ P пѓ§1 пЂ« пѓ· nпѓё пѓЁ nt Where пѓ�A is the amount in the account at time, t пѓ�P is the principle (initial) amount пѓ� r is the decimal value of the interest rate пѓ� n is how many times per year the interest is compounded. Look for terms like: daily (n =365), semi-annually (n = 2), weekly (n = 52) and monthly (n =12) INTEREST INTEREST Ex 2. An bank account earns interest compounded monthly. The investment doubles in 9.27 years. Calculate the annual interest rate. r пѓ¶ пѓ¦ A пЂЅ P пѓ§1 пЂ« пѓ· 12 пѓё пѓЁ r пѓ¶ пѓ¦ 2 P пЂЅ P пѓ§1 пЂ« пѓ· 12 пѓё пѓЁ r пѓ¶ пѓ¦ 2 пЂЅ пѓ§1 пЂ« пѓ· 12 пѓё пѓЁ 12 t When the money doubles there will be 2P in the account. So A = 2P 12 ( 9 . 27 ) 111 . 25 We are now solving for the base. We must вЂњundoвЂќ the exponent. Raise both sides to the 1/111.25. INTEREST INTEREST Ex 2. An bank account earns interest compounded monthly. The investment doubles in 9.27 years. Calculate the annual interest rate. r пѓ¶ пѓ¦ 2 пЂЅ пѓ§1 пЂ« пѓ· 12 пѓё пѓЁ 111 . 25 1 1 1 2 111 . 25 111 . 25 111 . 25 пѓ©пѓ¦ пѓ№ r пѓ¶ пЂЅ пѓЄпѓ§ 1 пЂ« пѓ· пѓє 12 пѓё пѓЄпѓ« пѓЁ пѓєпѓ» 1 2 111 . 25 пЂЅ 1пЂ« r 12 2 111 . 25 пЂ1 пЂЅ r 12 1 пѓ¦ пѓ¶ 111 . 25 12 пѓ§ 2 пЂ 1пѓ· пЂЅ r пѓ§ пѓ· пѓЁ пѓё r пЂЅ 0 . 075 r пЂЅ 7 .5 % INTEREST INTEREST Ex 3. Which is better: 5% interest per year compounded monthly or 5% per year compounded daily? LetвЂ™s assume an initial investment of $100 and a term of 10 years. 0 . 05 пѓ¶ пѓ¦ A пЂЅ 100 пѓ§ 1 пЂ« пѓ· 12 пѓЁ пѓё A пЂЅ 164 . 70 12 (10 ) 0 . 05 пѓ¶ пѓ¦ A пЂЅ 100 пѓ§ 1 пЂ« пѓ· 365 пѓё пѓЁ 365 (10 ) A пЂЅ 164 . 87 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.

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