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Square roots Square root the top and the bottom separately. Leave answer as a fraction – not a decimal. Yes, you can do this in your head! 0.4 0.4 0.16 is like knowing 4 4 16 and moving the decimal two places left. 25 25 5 36 36 6 0.16 0.4 Simplify nth roots 3 320 3 64 5 3 64 3 5 4 3 5 Another example: 3 375 3 125 3 3 125 3 3 5 3 3 Find a number in the X3 column on your yellow sheet that goes into 320 evenly (in this case 64). 64 is a “perfect cube” because it can be written as a product of whole numbers (in this case 64=4x4x4). 64 gets to come “outside” of the radical only because it’s “perfect”. When it comes outside, you write the root (in this case 4). Any numbers that aren’t “perfect” have to stay “inside” the radical. x 3 and y 3 are both perfect cubes because 3 3 x11 y 5 3 x3 x3 x3 x 2 y 3 y 2 xxxy 3 x 2 y 2 x 3 y 3 x 2 y 2 x3 x and 3 y3 y . These “perfect” cubes get to come “outside” of the radical. Break the x11 into multiples of x3 and then any remainder. The remainders are stuck “inside” the radical because they aren’t “perfect”. Note: on the multiple choice problems, make sure the answer you choose has a root that matches the one in the problem. In other words x 2 y 3 x 2 y1 not x 2 y x 2 y1 3 x8 y 4 8 3 2 with a remainder of 2 4 3 = 1 with a remainder of 1 x 2 y 3 x 2 y1 There’s a shortcut if you want to try it. Take the power on the x (in this case 8) and divide it by the root (in this case 3). 3 goes into 8 2 whole times with a remainder of 2. This means that the x on the outside has a power of 3 and the x on the inside has a power of 2. Similarly, the exponent on the y is 5 and 3 goes into it 1 whole time with 2 left over. Simplify expressions with rational exponents x 1 4 x 4 5 x 5 20 x 16 20 x 21 20 1 x 21 20 Another example: x 1 2 x 2 3 x 3 6 x 4 6 x 7 6 16 x 1 3 6 x x x 1 3( 6) 16 1 318 3 1 6 x17 3 16 Alternatively: 16 x 1 3 6 x x 1 36) 16 x17 18 x 1 3 1 6 x 6 1 x7 6 16 1 Convert both fractions to a denominator of 20 and then add them. The answer here had a negative exponent and you shouldn’t write answers with negative exponents so use the rule where you put the expression in the denominator and make the exponent positive. 1 am m a Remember to do the ( ) first. When you divide numbers with like bases, subtract the exponents. am a mn n a In this case the exponents on x are 1 6 and 6 or 3 1 Remember that when you subtract a negative, change it to adding a positive. The common denominator in this case is 3 so 18 multiply both the 6 and the 1 by 3 which is . 3 Add the exponents keeping the denominator of 3. Then use the power to a power rule and x 1 18 1 x x 1 18( 1) x 1 1818 18 x17 18 multiply the exponents. a m n a mn Write square roots as exponential expressions Express using fractional exponents: g g 2 g1 g1 2 Another example: Write 12z in exponential form. Remember that when we write a square root, we usually don’t write the little 2 but it’s there. Similarly, the g has a power of 1 even though we don’t usually write it. The power (1) is the top of the fraction and the root (2) is the bottom of the fraction. 12 z 2 12 z 12 z 1 12 Express using fractional exponents 6 6 j 5 The power (5) is the top of the fraction and the root (6) is the bottom of the fraction. j5 j5 6 Page 1 Simplify expressions with fractional exponents x m Simplify: 8 2 2 3 1 8 23 x n m 3 8 253 2 2 253 2 15625 125 but I prefer to take the root first and then use the exponent. First rewrite it with a positive exponent, then find the cube root of 8 (which is 2) and then take that answer to the power of 2. 5 5 5 5 125 3 25 m x n or The bottom of the fraction is the root. 2 2 2 8 so 3 8 equals 2 The bottom of the fraction is the root and the top is the power. You could also write it as 81 3 3 8 2 253 2 n 1 3 2 1 2 2 1 4 Solve equations containing radicals Solve: x 5 6 Solve: x 2 3 8 There is no way for the square root of a negative number to equal a negative number so the answer to this problem is “no solution”. This is similar to the previous problem, but you need to add 3 to both sides first and then realize that you can’t square root any number to get a negative. x 2 3 8 +3 x 2 5 4 x 2 3x 4 Solve: +3 4x 2 2 3x 4 4 x 2 3x 4 3x -3x 2 1. Square both sides. 2. Put the x ’s on the left side by adding 3x to both sides. 3. Put the numbers on the right by adding 2 to both sides. 1x 2 4 +2 Solve: 3 3 +2 x 2 x 6 5 x6 3 5 3 x 6 125 +6 +6 This is different from the square root problems because you CAN take the cube root of a number and get a negative answer. 1. Cube both sides 2. Add 6 to both sides x 119 Page 2 Solve exponential equations 1 273 x 4 9 2 3 33(3 x 4) Solve: 2 9 x 12 12 Get a common base on both sides of the equation, in this case use 3. Once you have this, you can just use the exponents because they have to be equal if the bases are equal. Then SOLV4X! +12 10 9 x 10 9 Solve: 5.87 x 24 x log 5.87 log 24 x log 24 log 5.87 x 1.796 Find the solution: 3x 4 x8 3 x 4 x 48 x log 3 x log 4 8log 4 x log 3 x log 4 8log 4 x(log 3 log 4) 8log 4 8log 4 x log 3 log 4 x 38.5507 x Take the log of both sides and then treat the log5.87 as a constant (which it is) and divide both sides by it. The answer rounds to 1.796 or 1.8. The sign means “is approximately equal to” and should be used to show that you rounded an answer. You can’t get a common base using 3 and 4. I can’t find an easier way to do this problem although there are other ways to do it. You can evaluate the logs sooner but then you might have a rounding error that you don’t get if you wait until the last minute to use your calculator. Page 3 WP: Roots (word problems) A certain gas will escape from a storage tank according to the formula e 130 p1 2 , where e represents the amount escaping in gallons per minute, and p represents the pressure in pounds per square inch. How much gas is escaping when the pressure is 1600 pounds per square inch? Round your answer to the nearest whole number. e 130(1600)1 2 You are given the pressure ( p 1600 ), so substitute 1600 for p . Take the square root of 1600 to get 40 and then multiply 40 by 130. The answer is expressed in gallons per minute. e 130 40 e 5200 gal/min The volume of a sphere can be given by the formula V 4.18879r 3 . You have to design a spherical container that will hold a volume of 65 cubic inches. What should the radius of your container be? 65 4.18879r 3 65 r3 4.18879 You are given the volume needed ( V 65 ), so substitute 65 for V . To get r 3 by itself, divide both sides by 4.18879 and then cube root both sides to get just r . You can simplify the left side using you calculator in one step: 3 (65 / 4.18879) or 3 x (65 / 4.18879) . Both the 3 m. 65 r 4.18879 r 2.49 inches 3 Page 4 and the can be found by pressing Find Inverses of functions Find the inverse of f ( x) 7 x 2 . Determine if the inverse is a function. x 7 y2 x y2 7 x y 7 f 1 ( x ) f ( x) is the same as y, so the equation is really y 7 x 2 . To find the inverse, switch the x and y and solve for y. x 7 Write the equation for the inverse f ( x) y 5 8 8x y 5 x 5 8 x To determine whether or not it’s a function you can graph it. Remember to graph both x x and y . This will make a y 7 7 parabola that opens to the right and if you apply the vertical line test, it proves that it is not a function. f ( x) is the same as y, so the equation is y 5 really x . To find the inverse, switch 8 the x and y and solve for y. This is a linear equation so it will be a function. y 8x 5 f 1 ( x ) 8 x 5 Exponential equations to log form b x a can be written as logb a x 2 27 is the base, 2 Write the equation 27 3 9 in logarithmic form. log 27 9 2 3 3 is the exponent and 9 is the answer. Log equations to exponential form logb a x can be written as b x a Write the equation log 243 9 form. 2 in exponential 5 243 is the base, 2 the answer. 2 243 5 9 Page 5 5 is the exponent and 9 is Find logs by converting to exponential form b x a can be written as logb a x baseexponent answer can be written as logbase answer exponent Evaluate log10 10 5 log10 105 x 10 x 105 answer: -5 Evaluate log5 25 log 5 25 x 5x 25 5 x 52 answer: 2 1 Evaluate log 3 729 1 log 3 x 729 1 3x 729 1 3x 6 3 x 3 36 answer: -6 Convert to exponential form and SOLV4X. Since both sides have a base of 10, then x 5 . A shortcut would be to just look at the exponent on the 10. Don’t make it harder than needed! Convert to exponential form and SOLV4X. The exponent that you have to put on 5 to get 25 is 2 so the answer is x 2 . Convert to exponential form and SOLV4X. The exponent that you have to put on 3 to get 729 is 6, but it’s in the bottom of a fraction so the exponent has to be -6. Page 6 Find natural logs of numbers ln b is read “natural log of b” and is a simplified way of writing loge b . Do these types of problems by using e as the base. There is an e function on your calculator above the / button. Evaluate ln 34 ln 34 3.526 log e 34 3.526 Calculator: L 34 e. Convert to exponential form. e3.526 34 1 Evaluate ln 8 e 1 ln 1 e 8 log e x ln e 1 Convert from radical form to a fractional exponent. You can skip the last few steps if you recognize that you are looking for the exponent on e . 8 1 8 x just look at the When you get to log e e e e e x 1 x 1 exponent on the e . 8 8 Page 7 1 8 Properties of logs, simplify expressions log a xy log a x log a y log a Express in terms of logarithms of x, y and z: 3 xy 4 log a 2 z log a 3 xy 4 log z 2 x log a x log a y y log a x m m log a x Use the rules for logarithms of products, quotients and powers to write as the sum or difference of logarithms: log b log a 3 log a x log a y 4 log z 2 9 log b x log a 3 log a x 4 log a y 2 log z Given logc 2 0.315 , log c 3 0.500 and 5 logc 5 0.732 , find log c . 12 5 log c 2 2 3 log c 5 2 log 2 log 3 x7 y3 x7 y3 log b 8 z8 z 7 9 log b y 1 3 log b z 1 9 7 log b x 9y z 8 9 8 3 9 9 7 1 8 log b x log b y log z 9 3 9 Write as a single logarithm: 4logb x 8logb y log b x 4 log b y 8 log b x 4 log b y 8 0.732 2 0.315 0.500 .398 Properties of logs, find equivalent logs log a xy log a x log a y log a x log a x log a y y 64 Find an expression equivalent to log a 1 3 x x 3 64 log a 3 3 x x log a x m m log a x Find an expression equivalent to log a x 3 64 log a 3 x ( x 4)( x 2 4 x 16) log a x3 2 log a ( x 4) log a ( x 4 x 16) 3log a x This is a really hard problem because you have to get a common denominator so you can subtract the fractions and then know how Page 8 1 61 2 log a 64 1 log a 61 log a 64 2 61 64 to factor x3 64 into ( x 4)( x 2 4 x 16) . Solve log equations Solve: log2 x 8 2 Convert to exponential form. 2 2 x 8 1 x 8 4 8 8 1 32 4 4 31 x 4 Solve: log3 x 6 log3 x 4 SOLV4X Subtract 8 from both sides. Convert -8 into a fraction. Subtract the fractions. x log 3 x 6 4 1x 1x x x 6 34 x 81x x 6 Combine the logs. Convert to exponential form. SOLV4X Multiply both sides by x . Subtract 1x from both sides. Divide both sides by 80. Reduce the fraction. 80 x 6 6 80 3 x 40 x Solve equations involving natural logs Find x if e9.2 x 6 and you are given ln 6 1.7918 log e 6 9.2 x Convert to log form. Because loge x ln x , you can substitute 1.7918 for ln 6 and divide by 9.2. ln 6 9.2 x ln 6 9.2 x .1948 x Page 9 Graph exponential and log functions Graph f ( x) 2 x 5 Graph y log 2 x 3 Graph f ( x) 2 x on your calculator. The equation is shifted 5 units down from this parent equation. You can’t graph y log 2 x on your calculator because it only has log and ln functions. Instead, use this conversion: log x logb x log b In the case of this example, that would be y log x / log 2 3 . What do you think the -3 does to the graph? Graph y log( x 1) / log 2 on your calculator. Graph y log 2 ( x 1) What is the difference between this problem and the previous one. Hint: the -3 was outside the ( ) in the last one and the -1 is inside the ( ) on this one. This makes a difference in how the graph is shifted. Page 10 WP: Interest Problems The compound amount with continuous compounding is given by A Pe rt , where P is the principal, r the rate and t the time in years. If the rate is 9.5%, find how long it takes for the money to double, that is for A 2 P . ln 2 0.69315 2 P Pe.095t Substitute 2P for A . Divide both sides by P . Convert to ln form. Divide both sides by .095. Substitute 0.69315 for ln2. It will take 7.3 years for the money to double. 2 e.095t ln 2 .095t ln 2 .095 0.69315 t .095 t 7.3 years t If $10,000 is invested at 6% compounded monthly for 7 years the compounded amount 84 is given by A 10, 000 1.005 . Given that log1.005 0.00217 , find log A . (Note that Note that you are asked to find log A not just A . Convert to log form and simplify. 10, 000 104 ) log A log104 84 log1.005 log A 4 84 0.00217 log A 4.18228 Page 11 WP: Growth & decay problems A radioactive substance decays so that the amount A present at time t (years) is A A0 e 2.5t . Find the half-life (time for half to decay) of this substance. .5 A0 A0 e ln.5 0.69315 A0 is the “initial amount” of the substance. Substitute 0.5A for A0 because you want the half-life of the initial amount of the substance. 2.5t Divide both sides by A0 . .5 e 2.5t ln .5 2.5t Convert to ln form. ln 0.5 2.5 0.69315 t 2.5 t 0.277 years The number of a certain product that will be sold t years after the product is introduced is given by 7300ln 5t 2 . How many of the t product will be sold 7 years after the product is introduced? Round the answer to the nearest whole number. 7300 ln 5t 2 Solve for t . Substitute 7 for t. Simplify using a calculator. Round 26,359.7 to 26.360. 7300 ln 5 7 2 7300 ln 37 26,360 Page 12