# Thread: Graphs and equations of straight line

1. ## Graphs and equations of straight line

The variable x and y are connected by the law y=Ce^ax,where a,C and e are constant.The experimental values of x and y are recorded in the following table

x 1 2 3 4
y 6.59 10.87 17.90 29.56

plot a graph of log y against x.Given that log e=0.4343,use graph to find the values of a and c..

2. Originally Posted by mastermin346
The variable x and y are connected by the law y=Ce^ax,where a,C and e are constant.The experimental values of x and y are recorded in the following table

x 1 2 3 4
y 6.59 10.87 17.90 29.56

plot a graph of log y against x.Given that log e=0.4343,use graph to find the values of a and c..

The first step is to take the logarithm of y and plot those points. If $y= Ce^{ax}$ then $log(y)= log(C)+ a log(e)x$, a straight line with slope a log(e) and y-intercept log(C). (The logarithm here is the common logarithm, base 10, since they tell you that log(e)= 0.4343)
The y value, where the line crosses the y-axis, is log(C) and so $C= 10^y$ for that y value. a is the slope of that line divided by log(e)= 0.4343. If this were exactly a straight line, you could calculate the slope using any two points but with inexact measurements, it is best to use the two points farthest apart: the slope would be $\frac{log(29.56)- log(6.59)}{4- 1}$.
The first step is to take the logarithm of y and plot those points. If $y= Ce^{ax}$ then $log(y)= log(C)+ a log(e)x$, a straight line with slope a log(e) and y-intercept log(C). (The logarithm here is the common logarithm, base 10, since they tell you that log(e)= 0.4343)
The y value, where the line crosses the y-axis, is log(C) and so $C= 10^y$ for that y value. a is the slope of that line divided by log(e)= 0.4343. If this were exactly a straight line, you could calculate the slope using any two points but with inexact measurements, it is best to use the two points farthest apart: the slope would be $\frac{log(29.56)- log(6.59)/(4- 1)$.
4. Sorry, I messed up the LaTex exactly where I was addressing that point. The slope of the line is $log(29.56)- log(6.59)}{4- 1}$ and a is that number divided by 0.4343.