Logarithmic Functions


 Eustacia Manning
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1 Logarithmic Functions = a b + c + d = a b + c + d log ln n Logarithmic unctions will be graphed in the same manner as radical unctions. It is irst necessary to ind the domain o the logarithmic unction. The range o a logarithmic unction is all real numbers, so only the domain needs to be ound. To ind the domain o a logarithmic unction evaluate b + c > 0. Remember, this is not, because you cannot take the log o zero. Once the domain is ound, it will tell in which direction the unction is moving. This inequality will also help ind the vertical asymptote or the unction. I the inside the log does not have a negative coeicient, the curve will be on the right side o the vertical asymptote. I the coeicient in ront o is 1, begin with the key point o (1,0). From that point on, treat the unction just like an eponential unction. Adding or subtracting to either the or y values to ind the new key point making the graph shit. I the inside the log has a negative coeicient, the curve will be on the let side o the vertical asymptote. I the coeicient in ront o is 1, begin with the key point o (1,0) and shit rom there. *Once again, just like eponential growth and decay unctions, watch the value o a, as it aects the scale o the unction. I the value o a is some number other that 1 or 1, ind the key point algebraically beore you translate the unction. As the unction shits, it will be helpul to draw a broken line or both the horizontal and vertical asymptotes. It is OK to cross the horizontal asymptote, as you will ind the key point always rests on it. The vertical asymptote, however, may never be crossed. = a b + c + d = a b + c + d log ln n Solving or b + c = 0, will yield the equation or the vertical asymptote. The equation or the horizontal asymptote is y = d. Finding the domain. 3 ( ) = log > 0 Notice the similarity in the procedures. > 4 Finding the vertical asymptote. 4 = 0 = 4 *I the variable inside the log has a coeicient other than 1 or 1, the key point will be dierent. The key point must then be ound algebraically. To ind the value o the key point solve or b + c = 1. Substitute that solution back into the problem to ind the y value. Finding the horizontal asymptote. y = There is no real work involved with inding the horizontal asymptote. Identiy the vertical shit. This is the equation o the horizontal asymptote.
2 = a b + c + d = a b + c + d log ln n = log The parent unction has the key point at (1, 0) = log ( 3 ) = log ( + 3 ) The graph o this unction shits right 3. Notice the key point moved to the right 3 places to (4,0). The graph o this unction shits to the let 3. The new key point is (,0). = log + = log This unction shits up. Add to the y value o the key point, and it is now at (1,). This unction shits down. Subtracting rom the y value o the key point results in (1,).
3 = a b + c + d = a b + c + d log ln n = log log ( ) The parent unction has the key point at (1, 0) = = log The graph o this unction relects about the vertical asymptote. Key point is now (1,0). The graph o this unction is relected about the horizontal asymptote. Key point is still at (1,0). log ( ) = = log Since the coeicient o is 1, this graph will be on the let side o the vertical asymptote. Begin with the key point (1,0), and shit right because it is positive. Add to the value o the key point. The new key point is (1,0). Notice the negative portion o the graph relected above the ais. Match the appropriate graph with its equation below. Eplain why each o your solutions is
4 true. A B C D E F 1) = log ( ) ) log ( 1 ) = 3) = log ) = log3 + 5) = log 6) = 3log
5 The translation o a logarithmic unction is almost identical to that o an eponential unction. Just make sure to identiy on which side o the vertical asymptote the graph o the unction will reside. This will determine which key point to begin with. Remember to draw both asymptotes to graph the unction and watch or the value o a which will aect key point. Graph each o the ollowing logarithmic unctions by inding the asymptotes and labeling the key point. Be sure to ind the intercept and y intercept (i they eist). A) = log3 + B) = log ( + 3)
6 C) = log + 4 D) = ln ( )
7 E) = ln ( ) F) = log 3 +
8 G) = ln H) = ln
9 All standard logarithmic unctions (meaning a unction without absolute value symbols), must have an intercept. All standard eponential growth and decay unctions must have a y intercept. Are these two statements true? Why or why not? In order to ind the domain o the logarithmic unction ( ) evaluate + 5 > 0. Why must we use this inequality? = log + 5 3, we need to 4 What is the problem with relying on a graphing calculator to graph a logarithmic unction?
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