course mth 158 As I was reading through the chapter, I understood each section (with maybe some difficulty in the word problems) but I felt that I had a really good understanding of the chapter until I tried to solve the problems here. I don't know how to improve my skills on solving problems, but I keep trying. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). -3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate. To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I still really don’t understand this one.
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Given Solution: * * If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2. Since e^0 = 1 this tells us that A0 = 100 mm^2. So the function is A(n) = 100 mm^2 * e^(-.35 n). To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx.. After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm63 * .0302 = 3.02 mm^2 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t think there are any discrepancies. I can’t believe I solved this word problem without help!!
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Given Solution: * * * 7th edition only 5.4.14. 2.2^3 = N. Express in logarithmic notation. Read as 3 is the logarithm at base 2.2 of N expressed in notation would be 3 = log{base 2.2}N b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 2.2, x = 3 and y = N. So we write lob{base b}(y) = x as log{base 2.2}(N) = 3. 8th edition only 6.4.14. e^2.2 = N. Express in logarithmic notation. b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 3, x = 2.2 and y = N. So we write lob{base b}(y) = x as log{base e}(N) = 2.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I wrote it backwards. I have a good basic understanding of the logarithmic function so far. (I hope) ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * extra problem / 7th edition 5.4.18. x^pi = 3. Express in logarithmic notation. pi is the logarithm at base x of 3 log{base x}(3) = pi b^a = y is expressed in logarithmic notation as log{base b}(y) = a. In this case b = x, a = pi and y = 3. So we write lob{base b}(y) = a as log{base x}(3) = pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * extra problem / 7th edition 5.4.26.. log{base 2}?? = x. Express in exponential notation. x is the logarithm of base 2 of y or 2^x = y log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 2, a = x and y = ?? so the expression b^a = y is written as 2^x = ??. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not sure if I should have put the y in my equation. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 6.4.30 / 7th edition 5.4.36. Exact value of log{base 1/3}(9) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve log{base 1/3}(9) you set y = log{base 1/3}(9) y = log{base 1/3}(9) 1/3^y = 9 1/3^y = 1/3^-2 we know that 3^2 = 9 so the reciprocal of 2 which is -2 should work here y = -2 Therefore, log{base 1/3}(9) = -2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 1/3, a is what we are trying to find and y = 9 so the expression b^a = y is written as (1/3)^a = 9. Noting that 9 is an integer power of 3 we expect that a will be an integer power of 1/3. Since 9 = 3^2 we might try (1/3)^2 = 9, but this doesn't work since (1/3)^2 = 1/9, not 9. We can correct this by using the -2 power, which is the reciprocal of the +2 power: (1/3)^-2 = 1/ ( 1/3)^2 = 1 / (1/9) = 9. So log base 1/3}(9) = -2. . What is the domain of G(x) = log{base 1 / 2}(x^2-1) For any positive value of b the domain of log{base b}(x) consists of all positive real numbers. It follows that log{base 1/2}(x^2-1) consists of all real numbers for which x^2 - 1 > 0. We solve x^2 - 1 > 0 for x by first finding the values of x for which x^2 - 1 = 0. We can factor the expression to get (x-1)(x+1) = 0, which is so if x-1 = 0 or if x + 1 = 0. Our solutions are therefore x = 1 and x = -1. It follows that x^2 - 1 is either positive or negative on each of the intervals (-infinity, -1), (-1, 1) and (1, infinity). We can determine which by substituting any value from each interval into x^2 - 1. On the interval (-1, 1) we can just choose x = 0, which substituted into x^2 - 1 gives us -1. We conclude that x^2 - 1 < 0 on this interval. On the interval (-infinity, -1) we could substitute x = -2, giving us x^2 - 1 = 4 - 1 = 3, which is > 0. So x^2 - 1 > 0 on (-infinity, -1). Substituting x = 2 to test the interval (1, infinity) we obtain the same result so that x^2 - 1 > 0 on (1, infinity). We conclude that the domain of this function is (-infinity, -1) U (1, infinity). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t see any discrepancies, but I’m not sure where this came from: . What is the domain of G(x) = log{base 1 / 2}(x^2-1)
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Given Solution: * * We easily solve for a by taking the -1/4 power of both sides, obtaining a = (1/2)^-(1/4) = 16. . Transformations to graph h(x) = ln(4-x). Given domain, range, asymptotes.lineCount = lineCount + 1: bLine$(lineCount) = "" The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axilineCount = lineCount + 1: bLine$(lineCount) = s and passes through the points (1, 0) and (e, 1) (the latter is approximately (2.71828, 1) ). The graph of y = ln(x - 4), where x is replaced by x - 4, is shifted +4 units in the x direction so the vertical asymptote shifts 4 units right to the line x = 4. The points (1, 0) and (e, 1) shift to (1+4, 0) = (5, 0) and (e + 4, 1) (the latter being approximately (3.71828, 1). Since x - 4 = - (4 - x), the graph of y = ln(4 - x) is 'flipped' about the y axis relative to ln(x-4), so it the vertical asymptote becomes x = -4 and the graph passes through the points (-5, 0) and (-e-4, 1). The graph will still be concave down. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I couldn’t follow along with the solution on this one. It was a little confusing. Also, I think some of your code is showing up in the solution: it may be missing a ( ) or “ ” asymptotes.lineCount = lineCount + 1: bLine$(lineCount) = "" ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.4.98 / 7th edition 5.4.102. Solve log{base 6}(36) = 5x + 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is one that I tried all evening to solve. It looks like it is 5x+3 is the logarithm of base 6 at 36 so would you set it up like this: 6^5x+3 = 36 we know that 6^2 = 36 so 5x+3 has to equal 2? 5x+3 = 2 5x = 2-3 5x = 1 x = 1/5 6^5(1/5) + 3 = 1296 so that’s not right either. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 6, a is what we are trying to find and y = 5x + 3 so the expression b^a = y is written as 6^(5x+3) = 36. We know that 6^2 = 36, so (5x + 3) = 2. We easily solve this equation to get x = -1 / 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was on the right track. But how did you get -1/5?
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Given Solution: * * The probability is F(t), so F(t) = 50% = .5 when 1 - e^(-.15 t) = .5. We subtract 1 from both sides the multiply by -1 to rearrange this equation to the form e^(-.15 t) = .5. Taking the natural log of both sides we get ln( e^(-.15 t) ) = ln(.5), which tells us that -.15 t = ln(.5) so that t = -ln(.5)/.15 = 4.6, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I tried doing what the given solution suggests and subtracted 1 from both sides, but mine was -e^(-.5t) = -1/2 so I must have done something wrong. Then, when I tried to do it step by step with the given solution, I entered ln(e^-.15) and came up with -0.15 and ln(.5) = -0.69 so how does that equal. I am not clear on how to solve this equation.