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Use Formula 10 to graph the given functions on a common screen. How are these graphs related?
$ y = \ln x $ , $ y = \log_{10} x $ , $ y = e^x $ , $ y = 10^x $
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00:52
Jeffrey Payo
Calculus 1 / AB
Calculus 2 / BC
Calculus 3
Chapter 1
Functions and Models
Section 5
Inverse Functions and Logarithms
Functions
Integration Techniques
Partial Derivatives
Functions of Several Variables
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in this problem. We're using a graphing calculator and we're graphing four grafs on a common screen, and we're seeing how they're related. So we go to y equals and we type them in. We have y equals natural log X Y equals log based 10 of X y equals E to the X and y equals 10 to the X. And for my screen, I decided to go with negative 6 to 10 on the X axis and negative 6 to 10 on the Y axis. You can fiddle around with these numbers and change them to get something that you like. So here, all four graphs at the same time. And if I press trace, I can tell which is which. So the blue one is the natural law graph, and the red one is log based 10. So these have the same basic shape. They both have the same X intercept at one, but the one with a greater base is the lower graph. It's a bit more flat. The black graph is each of the X, and the pink one is 10 to the X, so they have the same basic shape. They're both exponential growth but the one with a greater base grew a lot faster. Tend to the X grows a lot faster than E to the X, and the other thing we want to note is that we do have some inverse is going on here. So the pink one y equals 10th E X is the inverse of the red one, so you can see that pink and red look like reflections across the line Y equals X. In fact, if I graph the line Y equals X, that might help. It's not going to look like it's out of 45 degree angle because my window is kind of skewed. But still, we can see that as a reflection line between the pink and the red. And also the blue and the black are in verses. The blue one was the natural log function, and the black is the E to the X function, and they're also reflections across the line Y equals X
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