#$&* course mth 158 8/7/13 2:41 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 2.2, x = 3 and y = N. So we write lob{base b}(y) = x as log{base 2.2}(N) = 3. 8th edition only 6.4.14. e^2.2 = N. Express in logarithmic notation. b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 3, x = 2.2 and y = N. So we write lob{base b}(y) = x as log{base e}(N) = 2.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * extra problem / 7th edition 5.4.18. x^pi = 3. Express in logarithmic notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^pi = 3 log {base x} 3 = pi confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * extra problem / 7th edition 5.4.26.. log{base 2}?? = x. Express the equation log{base 2} ( A ) = x in exponential notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 2} ( A ) = x A = 2^x confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 2, a = x and y = A so the expression b^a = y is written as 2^x = A. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 6.4.30 / 7th edition 5.4.36. Exact value of log{base 1/3}(9) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 1/3}(9) 9 = (1/3)^x -2 = x confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question:. What is the domain of G(x) = log{base 1 / 2}(x^2-1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I really was not sure how to approach this one. I wanted to say it would be all real numbers at first, but I was not totally sure how to work it to find out for sure. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For any positive value of b the domain of log{base b}(x) consists of all positive real numbers. It follows that log{base 1/2}(x^2-1) consists of all real numbers for which x^2 - 1 > 0. We solve x^2 - 1 > 0 for x by first finding the values of x for which x^2 - 1 = 0. We can factor the expression to get (x-1)(x+1) = 0, which is so if x-1 = 0 or if x + 1 = 0. Our solutions are therefore x = 1 and x = -1. It follows that x^2 - 1 is either positive or negative on each of the intervals (-infinity, -1), (-1, 1) and (1, infinity). We can determine which by substituting any value from each interval into x^2 - 1. On the interval (-1, 1) we can just choose x = 0, which substituted into x^2 - 1 gives us -1. We conclude that x^2 - 1 < 0 on this interval. On the interval (-infinity, -1) we could substitute x = -2, giving us x^2 - 1 = 4 - 1 = 3, which is > 0. So x^2 - 1 > 0 on (-infinity, -1). Substituting x = 2 to test the interval (1, infinity) we obtain the same result so that x^2 - 1 > 0 on (1, infinity). We conclude that the domain of this function is (-infinity, -1) U (1, infinity). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This makes more sense finding the intervals and testing them to find the domain. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 6.4.58 / 7th edition 5.4.62. a such that graph of log{base a}(x) contains (1/2, -4). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base a}(x) converted into a function f (x) = log{base a}(x) if x = 1/2 and y = -4 then -4 = log { base a} (1/2) 1/2 = a ^-4 1/2 = 1/a^4 a^4/2 = 1 a^4 = 2 a = 2^1/4 a approx= 1.189 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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: (a ^ (-4))^(-1/4) = 1/2 ^ (-1/4) so since (a^(-4))^(-1/4)) = a^(-4 * (-1/4)) = a^1 = a, and 1/2 ^(-1/4) = 2^(1/4) = sqrt( sqrt(2) ), or approximately 1.19. a = (1/2)^-(1/4) = 2^(1/4) = sqrt( sqrt(x) ), which is approximately 1.19. So we have log{base sqrt(sqrt(x))} (1/2) = -4 since (sqrt(sqrt(x)))^ (-4) = 1/2. Note also that 1.19^(-4) is .499, very close to 1/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Use transformations to graph h(x) = ln(4-x). Give domain, range, asymptotes. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axis and passes through the points (1, 0) and (e, 1) (the latter is approximately (2.71828, 1) ). The graph of y = ln(x - 4), where x is replaced by x - 4, is shifted +4 units in the x direction so the vertical asymptote shifts 4 units right to the line x = 4. The points (1, 0) and (e, 1) shift to (1+4, 0) = (5, 0) and (e + 4, 1) (the latter being approximately (3.71828, 1). Since x - 4 = - (4 - x), the graph of y = ln(4 - x) is 'flipped' about the y axis relative to ln(x-4), so it the vertical asymptote becomes x = -4 and the graph passes through the points (-5, 0) and (-e-4, 1). The graph will still be concave down. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.4.98 / 7th edition 5.4.102. Solve log{base 6}(36) = 5x + 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 6}(36) = 5x + 3 6^(5x+3) = 36 6^ (5x + 3) = 6 ^2 5x + 3 = 2 5x = -1 x = -1/5 confidence rating #$&*: 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 = 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 6.4.122 / 7th edition 6.3.102. F(t) = 1 - e^(-.15 t) prob of car arriving within t minute of 5:00. How long does it take for the probability to reach 50%? How long to reach 80%? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a) 50% = f f (t) = 1 - e^(-.15t) .5 = 1 - e^(-.15t) -.5 = -e^(-.15t) .5 = e ^(-.15t) -.15t = ln.5 t = ln.5 / -15 t approx = 4.621 min b) 80% = f .8 = 1 -e^(-.15t) -.2 = -e^(-.15t) .2 = e^(-.15t) -.15t = ln.2 t = ln.2 / .15 t approx = 10.730 min confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!