course Mth158 I am still trying very hard, but there is a wall between Chapters 4, 5 and me.
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17:32:09 5.4.14. 2.2^3 = N. Express in logarithmic notation.
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RESPONSE --> b=2.2, x=3 y=N log(base 2.2)(N)=3
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17:32:18 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
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17:32:55 5.4.18.. x^pi = 3. Express in logarithmic notation.
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RESPONSE --> b=x, a=pi y=3 log (base x)(3)= pi
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17:32:59 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
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17:33:36 5.4.26.. log{base 2}?? = x. Express in exponential notation.
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RESPONSE --> b=2 a=x y= ?? 2^x=??
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17:33:42 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
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17:35:37 5.4.36. Exact value of log{base 1/3}(9)
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RESPONSE --> b= 1/3 a= not known y=9 b^a=y 1/3^(unknown)=9 a=-2
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17:35:45 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
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17:37:21 . What is the domain of G(x) = log{base 1 / 2}(x^2-1)
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RESPONSE --> I am going to say the domain of this function is all real numbers
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17:38:04 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 --> This is a bit more in depth than I expected. I usually have problems with domain, but I can continue to work on that.
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17:38:18 5.4.62. a such that graph of log{base a}(x) contains (1/2, -4).
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RESPONSE -->
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17:39:25 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 --> I messed up and hit enter. Sorry about that. I see that this is a simple problem, solved by taking the -1/4 power of each side to end up with a= 16.
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17:44:25 . Transformations to graph h(x) = ln(4-x). Given domain, range, asymptotes.""
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RESPONSE --> I don't have any idea how to do this.
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17:45:20 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.
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RESPONSE --> I still don't understand this that well, so I will reread this section and watch 5.4 on CD until I get it. If I still have trouble, I will email. Sorry.
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17:48:17 5.4.102. Solve log{base 6}(36) = 5x + 3.
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RESPONSE --> b=6 a= is not known y=5x+3 6^5x+3=36 I am not sure where to go from here.
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17:49:41 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, so I know 6^2 is 36, so 5x+3= 2 subtract 3 from both sides and divide by 5 to get x= -1/5. I see now!
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17:51:40 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%?
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RESPONSE --> I don't have any idea how to set this up. I think it would be something like F(t)=50% 1-e^(.15t)=.5 but I don't know for sure.
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17:53:46 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..
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RESPONSE --> Anytime I take the natural log of something I get in trouble. Ok, so I was going along the right track- then I needed to subtract one from both sides and multiply by -1 to get everything rearranged= e^(-.15t)=.5 Then take the nat. log. to end up with t=-ln.5/.15= approx 4.6 I need more work on this. This is a rough chapter.
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