6.1 Exponential and Logarithmic Functions
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1 Section 6.1 Eponential and Logarithmic Functions Eponential and Logarithmic Functions We start our review o eponential and logarithmic unctions with the deinition o an eponential unction. Deinition 1 An eponential unction is a unction o the orm (t) = b t, where b > 0 and b 1. b is called the base o the eponential unction. More generall, a unction o the orm (t) = Ab t, where b > 0, b 1, and A 0, is also reerred to as an eponential unction. In this case, the value o the unction when t = 0 is (0) = A, so A is the initial amount. In applications, ou will almost alwas encounter eponential unctions in the more general orm Ab t. Since eponential unctions are oten used to model processes that var with time, we usuall use the input variable t (although o course an variable can be used). Graphs o Eponential Functions Let s irst review the properties or the basic eponential unction b t irst, and then note the minor changes or the more general orm Ab t. For a working eample, let s use base b =. Plotting this unction ields the ollowing graph. 16 (t)= t Figure 1. t 4 Note several properties o the graph in Figure 1:
2 Chapter 6 a) Moving rom let to right, the curve rises, which means that the unction increases as t increases. In act, the unction increases rapidl or positive t. b) The graph lies above the t-ais, so the values o the unction are alwas positive. Thereore, the range o the unction is (0, ). c) The graph has a horizontal asmptote = 0 (the t-ais) on the let side. This means that the unction almost dies out (the values get closer and closer to 0) as t approaches. What about the graphs o other eponential unctions with dierent bases? As can be seen in Figure (a), the graph o 3 rises aster that or > 0, and dies out aster or < Figure. 3 () = 4 (a) (b) Comparing unctions 1 () =, () = 3, and Net, add in 3 () = 4. The result is shown in Figure (b). Again, increasing the size o the base to b = 4 results in a unction which rises even aster on the right and likewise dies out aster on the let. Now let s tr smaller values o the base b. First plot the graph o 1 () = (1/) Figure 3. Comparing unctions 1 () = (1/), () = (1/3), and 3 () = (1/4) This graph is much dierent. It rises rapidl to the let, and almost dies out on the right. Compare this with () = (1/3) and 3 () = (1/4). As the base gets smaller, the graph rises aster on the let, and dies out aster on the right.
3 Section 6.1 Eponential and Logarithmic Functions 3 Using relection properties, it s eas to understand the appearance o these last three graphs. Note that ( ) 1 = ( 1 ) =, (6.1) so it ollows that the graph o ( ) 1 is just a relection in the -ais o the graph o (see Figure 4). g 5 Figure 4. Comparing unctions () = and g() = (1/) = Thus, we seem to have two dierent tpes o graphs, and thereore two tpes o eponential unctions: one tpe is increasing, and the other decreasing. Our eperiments above, along with a little more eperimentation, should convince ou that b is increasing or b > 1, and decreasing or 0 < b < 1. The irst tpe o unctions are called eponential growth unctions, and the second tpe are eponential deca unctions. Properties o Eponential Growth Functions: () = b with b > 1 The domain is the set o all real numbers. Moving rom let to right, the graph rises, which means that the unction increases as increases. The unction increases rapidl or positive. The graph lies above the -ais, so the values o the unction are alwas positive. Thereore, the range is (0, ). The graph has a horizontal asmptote = 0 (the -ais) on the let side. This means that the unction almost dies out (the values get closer and closer to 0) as approaches.
4 4 Chapter 6 Properties o Eponential Deca Functions: () = b with 0 < b < 1 The domain is the set o all real numbers. Moving rom let to right, the graph alls, which means that the unction decreases as increases. The unction decreases rapidl or negative. The graph lies above the -ais, so the values o the unction are alwas positive. Thereore, the range is (0, ). The graph has a horizontal asmptote = 0 (the -ais) on the right side. This means that the unction almost dies out (the values get closer and closer to 0) as approaches. Eample 1 Plot the graph o the unction () = (1.5). Identi the range o the unction and the horizontal asmptote. Since the base 1.5 is larger than 1, this is an eponential growth unction. Thereore, its graph will have a shape similar to the graphs in Figure. The graph rises, there will be a horizontal asmptote = 0 on the let side, and the range o the unction is (0, ). The graph can then be plotted b hand b using this knowledge along with approimate values at =, 1, 0, 1,. See Figure 5. () = (1.5) (a) Figure 5. (b) Graph o () = (1.5) Logarithmic Functions Recall that or each eponential unction () = b (b > 0, b 1), the inverse unction 1 () is called the logarithm o to base b, and is denoted b log b (). The deining relationship is given in the ollowing deinition.
5 Section 6.1 Eponential and Logarithmic Functions 5 Deinition I b > 0 and b 1, then the logarithm o u to base b is deined b the relationship v = log b (u) u = b v. (6.) In order to understand the logarithm unction better, let s work through a ew simple eamples. Eample Compute log (8). Label the required value b v, so v = log (8). Then b (6.), using b = and u = 8, it ollows that v = 8, and thereore v = 3 (solving b inspection). In the last eample, note that log (8) = 3 is the eponent v such that v = 8. Thus, in general, one wa to interpret the deinition o the logarithm in (6.) is that log b (u) is the eponent v such that b v = u. In other words, the value o the logarithm is the eponent. Eample 3 Compute log 10 (10 000). Again, label the required value b v, so v = log 10 (10 000). B (6.), it ollows that 10 v = , and thereore v = 4. Note that here again we have ound the eponent v=4 that is needed or base 10 in order to get 10 v = Eample 4 Compute log 3 ( 1 9). ) v = log 3 ( 1 9 = 3 v = 1 9 b (6.) = v = since 3 = 1 9 Eample 5 Solve the equation log 5 () = 1. log 5 () = 1 = 5 1 = b (6.) = = 5
6 6 Chapter 6 Eample 6 Solve the equation log b (64) = 3 or b. log b (64) = 3 = b 3 = 64 b (6.) = b = 3 64 = 4 Eample 7 Solve the equation log 1/ () =. = log 1/ () = ( ) 1 = b (6.) = = 1 ( 1 ) = = 4 Common and Natural Logarithm There are two bases that are used on a regular basis: base 10 and base e. Both o these show up in real world applications and in more advanced work. Common Logarithm. log() and log 10 () are equivalent notations. Thus, we have the deining relationship v = log(u) u = 10 v. Natural Logarithm. ln() and log e () are equivalent notations. Thus, we have the deining relationship v = ln(u) u = e v. Graphs o Logarithmic Functions The graph o the eponential unction () = b or b > 1 is shown in Figure 6(a), along with its inverse logarithmic unction 1 () = log b (). These two graphs are relections across the line =. Similarl, the graph or 0 < b < 1 is shown in Figure 6(b).
7 Section 6.1 Eponential and Logarithmic Functions 7 = = 1 1 (a) b > 1 (b) 0 < b < 1 Figure 6. The graphs o () = b and 1 () = log b () are relections across the line =. Because domains and ranges o inverse unctions are interchanged, it ollows that Propert 1 Domain(log b ()) = (0, ) and Range(log b ()) = (, ). In particular, note that the logarithm o a negative number, as well as the logarithm o 0, are not deined. Two particular points on the graph o the logarithm are noteworth. Since b 0 = 1, it ollows that log b (1) = 0, and thereore the -intercept o the graph o log b () is (1, 0). Similarl, since b 1 = b, it ollows that log b (b) = 1, and thereore (b, 1) is on the graph. Propert log b (1) = 0 and log b (b) = 1 Finall, since the graph o b has a horizontal asmptote = 0, the graph o log b () must have a vertical asmptote = 0. This behavior is a consequence o the act that inputs and outputs o inverse unctions are interchanged, and can be observed in Figure 6. In the inal eample below, we ll appl a transormation to the logarithm and see how that aects the graph. Eample 8 Plot the graph o the unction () = log ( + 1).
8 8 Chapter 6 The graph o () = log ( + 1) will be the same as the graph o g() = log () shited one unit to the let. The graph o g is shown in??(b). The -intercept (1, 0) on the graph o g will be shited one unit to the let to (0, 0) on the graph o. Likewise, the vertical asmptote = 0 on the graph o g will be shited one unit to the let to the line = 1 on the graph o. The inal graph o is shown in Figure Figure 7. The graph o () = log ( + 1).
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