course MTH 158 pe¯assignment #037
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11:20:36 5.4.14. 2.2^3 = N. Express in logarithmic notation.
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RESPONSE --> y = log to the base a of x x = a^y so 3 = log to the base 2.2 of N I don't know how to type ""to the base"" confidence assessment: 3
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11:20:44 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.
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RESPONSE --> ok self critique assessment: 3
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11:21:08 5.4.18.. x^pi = 3. Express in logarithmic notation.
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RESPONSE --> pi = log{base x}(3) confidence assessment: 3
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11:21:12 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.
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RESPONSE --> ok self critique assessment: 3
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11:21:32 5.4.26.. log{base 2}?? = x. Express in exponential notation.
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RESPONSE --> I don't know what's up with the question marks, but my book has the problem log{base 2}(6) = x, so I'll go with that. 6 = 2^x confidence assessment: 3
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11:21:37 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 = ??.
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RESPONSE --> ok self critique assessment: 3
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11:22:14 5.4.36. Exact value of log{base 1/3}(9)
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RESPONSE --> 9 = 1/3 ^ y 3^2 = 3^-1y -y = 2 y = -2 confidence assessment: 3
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11:22:20 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.
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RESPONSE --> ok self critique assessment: 3
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11:22:49 . What is the domain of G(x) = log{base 1 / 2}(x^2-1)
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RESPONSE --> G(x) = log{base 1 / 2}(x^2-1) --> G(x) = log{base 1 / 2}(x^1), right? Or am I reading that wrong?
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11:23:11 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).
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RESPONSE --> Ok, so I did read it wrong. Was it my fault, or was it typed in the problem incorrectly? self critique assessment: 2
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11:24:41 5.4.62. a such that graph of log{base a}(x) contains (1/2, -4).
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RESPONSE --> a^-4 = 1/2 a^(-4 * -1/4) = (1/2)^-1/4 a = 2^4 a = 16 confidence assessment: 3
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11:24:46 log{base a}(x) = y if a^y = x. The point (1/2, -4) will lie on the graph if y = -4 when x = 1/2, so we are looking for a value of a such that a ^ (-4) = 1/2. We easily solve for a by taking the -1/4 power of both sides, obtaining a = (1/2)^-(1/4) = 16.
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RESPONSE --> ok self critique assessment: 3
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12:32:45 . Transformations to graph h(x) = ln(4-x). Given domain, range, asymptotes.lineCount = lineCount + 1: bLine$(lineCount) = ""
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RESPONSE --> The domain is x | x > 4 The range is th set of all real numbers The vertical asymptote is x = -4 confidence assessment: 2
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12:35:59 The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axilineCount = lineCount + 1: bLine$(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.
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RESPONSE --> ok self critique assessment: 3
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12:39:42 5.4.102. Solve log{base 6}(36) = 5x + 3.
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RESPONSE --> 36 = 6 ^(5x + 3) 6^2 = 6^(5x + 3) 2 = 5x + 3 5x = -1 x = - 1/5 confidence assessment: 3
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12:39:49 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.
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RESPONSE --> ok self critique assessment: 3
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