Remember that log (a) is only defined if a is positive. So log (y-2) is only defined if y-2>=0. ie y>=2. That's why the calculator says what it does. So there will be no part of your graph under or on the line y=2.If you give the Y-value a negative number, then the calculator will say "Not a number"
Can you post the exercise statement verbatim? In this graphing exercise, I'm not sure if y is the independent variable or the dependent variable. In the former, the y-axis is horizontal; in the latter, the x-axis is horizontal.... What am I doing wrong with my graphing???
log2(y - 2) is undefined (in the Real number system) for any value of y that's 2 or less.If you give the Y-value a negative number, then the calculator will say "Not a number"
Remember that log (a) is only defined if a is non-negative. So log (y-2) is only defined if y-2>0. ie y>2. That's why the calculator says what it does. So there will be no part of your graph under the line y=2.
You say: "What am I doing wrong with my graphing???"
If you are going to draw the graph by plotting points, you simply need a lot more points, and join them with a smooth curve not straight lines.
Try y= 2.5 y=2.1, y=2.01 for example.
You also say: "I am not going to put it in a form without the log." But that would be the best way to graph this question.
x=2∗log2(y−2)
2x=log2(y−2)
y−2=22x
y=22x+2
Can you graph that exponential function either
1. by plotting points as before
OR even better (2) by considering transformations of the graph of y=2x ?
Except (using your calculator) you didn't get the same answers.Besides you get the same answers anyway …
You can google for lessons, but the main part in solving the given equation for y is understanding how to switch between logarithmic and exponential form. That comes directly from the definition of logarithms. Logarithms are a special way of writing exponents.I don't know how to flip the variables around and end up with an exponential form ...
JeffM said:[MATH]f(x) = log_{10}(x)[/MATH].
[MATH]f(-\ 2)= 0.01[/MATH]
[MATH]f(-\ 1)= 0.1[/MATH]
[MATH]f(0) = 1.[/MATH]
[MATH]f(0.5) \approx 3.2.[/MATH]
[MATH]f(1) = 10.[/MATH]
[MATH]f(1.125) \approx 13.3.[/MATH]
This is exactly what a graphing calculator does except with many more points.
I don't understand "log2 times y", but I do have some final comments on this exercise. In the op, you had made it sound like this was a graphing exercise. But, it's not. They gave you the graph.… instead of looking at it as x=2*log2(y-2), just look at it as x=log2 times y …
To ironsheep. Do you really read log2(y) as "log base 2 times y "?instead of looking at it as x =2 log(y-2), just look at it as x = log subscript 2 times y. That would mean as you exchange---- 2 to the power of x = 2 (Y-2). Y equals (2 to the power of x) plus 4 and all of that is divided by 2. Just have to make the formula look simpler, which will allow me to get rid of the log