Help with logarithms and graphs

[shinto1]

1. Determine the equation y=f(x) for each of the following cases in simplest form. All of the plots are straight lines and all coordinates are indicated with abscissa first, ordinate second (x,y).

-log y versus log x on a rectilinear graph passes through (5,7) and (2,3)

-(y-2)^2 versus x on a log-log graph passes through (1,2) and (3,4)

-√y versus x on a semi-log graph passes through (3,2) and (6,4)



2. Homework Equations

-Power Equation : y = bx^m

-Linear Equation: y = mx +b

-Exponential Equation: y = be^mx or y = b10^mx (e is base 10)

-Point slope formula: y-y1 = m(x-x1)



3. For the first problem, I found the slope, m, which is 4/3 (or 1.333) and then I used the point slope formula to get y = 1.333x + 0.35. This equation is in a linear format but the thing that confuses me is the "log y versus log x". I used this same process for the other problems but I'm still confused by the "(y-2)^2 versus x on a log-log graph" and
√y versus x on a semi-log graph". I really don't understand how to implement that part of the question into the solution, so any help would be appreciated.

 

[LawrenceC]

Hint:

On a non-log graph, the coordinates are (10000,10000000) and (100,1000).

 

[NascentOxygen]

Hi shinto1. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

-log y versus log x on a rectilinear graph passes through (5,7) and (2,3)

I took that to be a minus sign ("–") in front of your log function, and didn't realize it wasn't until I got to the -√y plotted on semi-log paper!

[PLAIN]https://www.physicsforums.com/Nexus/statusicon/user_online.png (y-2)^2 versus x on a log-log graph passes through (1,2) and (3,4)

[PLAIN]https://www.physicsforums.com/Nexus/statusicon/user_online.png √y versus x on a semi-log graph passes through (3,2) and (6,4)

Do you understand what LawrenceC wrote about the first graph?

If a log-log plot is a straight line, then those same points plotted on an ordinary graph (rectilinear) will likewise be a straight line, but much expanded.