# Relation of the form y=ab^x graph and solving help !

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#1
Hello, so I have been practicing plotting graphs of relations in the form of y=kx^n and y=ab^x. This question concerns the latter situation.

An experiment is performed to test the relationship y=ab^x between two variables x and y.
Reduce the relationship to linear form and draw a suitable graph to verify the relationship. From the graph calculate the approximate values of a and b.
I have attached the table of values of the variables found.

I have attempted to solve this problem but would like some advice on my workings and any improvements/corrections to improve upon my solution.
I believe that a relation in the form y=ab^x becomes Y=mx=c where;
Y=log y
m=log b
c=log a
To solve take logarithms to base 10 of both sides:
y=ab^x
log y = loga +logb^x
log y = log a +xlog b

In this instance then plot log y against x and the gradient of the line of best fit will provide an estimate for log b, while the 'log b intercept' will offer an estimate for log a.
To investigate a relation of the form y=ab^x plot log y against x in a table and use these values to construct a graph.

(I have attached this table and graph).

The gradient I believe can be calculated as approximately:
0.9-0.05/7-0
0.85/7
0.1214285714
Thus, log b = 0.1214285714 and b = 10^ 0.1214285714
b=1.322600162= 1.3

The 'log y-intercept' is 0.05
so log a = 0.05 and a=10^0.05
a= 1.122018 =1.1

The proposed relation is therefore y = 1.1(1.3)^x

I would appreciate if someone was able to offer advice and methods to improve/correct my answer. Thank you very much to anyone in advance.
😊
0
6 months ago
#2
(Original post by LukeWatson4590)
Hello, so I have been practicing plotting graphs of relations in the form of y=kx^n and y=ab^x. This question concerns the latter situation.

An experiment is performed to test the relationship y=ab^x between two variables x and y.
Reduce the relationship to linear form and draw a suitable graph to verify the relationship. From the graph calculate the approximate values of a and b.
I have attached the table of values of the variables found.

I have attempted to solve this problem but would like some advice on my workings and any improvements/corrections to improve upon my solution.
I believe that a relation in the form y=ab^x becomes Y=mx=c where;
Y=log y
m=log b
c=log a
To solve take logarithms to base 10 of both sides:
y=ab^x
log y = loga +logb^x
log y = log a +xlog b

In this instance then plot log y against x and the gradient of the line of best fit will provide an estimate for log b, while the 'log b intercept' will offer an estimate for log a.
To investigate a relation of the form y=ab^x plot log y against x in a table and use these values to construct a graph.

(I have attached this table and graph).

The gradient I believe can be calculated as approximately:
0.9-0.05/7-0
0.85/7
0.1214285714
Thus, log b = 0.1214285714 and b = 10^ 0.1214285714
b=1.322600162= 1.3

The 'log y-intercept' is 0.05
so log a = 0.05 and a=10^0.05
a= 1.122018 =1.1

The proposed relation is therefore y = 1.1(1.3)^x

I would appreciate if someone was able to offer advice and methods to improve/correct my answer. Thank you very much to anyone in advance.
😊
Looks about right for the log stuff.
Not sure about what the table in the top right of your working is? When you log those y's you don't get the values in the table on the left.
0
#3
(Original post by mqb2766)
Looks about right for the log stuff.
Not sure about what the table in the top right of your working is? When you log those y's you don't get the values in the table on the left.
Thank you very much for oyur reply. Sorry to be unclear. The table on the left is the orignal table of values from the question of the values of varibales x and y found in the experiment. You are absolutely right, I thought I had done something silly but could not see why. I have log x instead of log y in the table !

In which case I will correct the table and graph hold on... 😁
0
#4
(Original post by mqb2766)
Looks about right for the log stuff.
Not sure about what the table in the top right of your working is? When you log those y's you don't get the values in the table on the left.
Thank you very much for your reply. Sorry to be unclear. The table on the left is the original table of values from the question of the values of variables x and y found in the experiment. You are absolutely right, I thought I had done something silly but could not see why. I have log x instead of log y in the table !
Hopefully correcting this I have attached an amended table and graph.
The gradient I believe can be calculated as approximately:
3.4-0.48/7.2-0
2.92/7.2
0.405555556
Thus, log b =0.405555556 and b = 10^ 0.405555556
b=2.544225233= 2.54

The 'log y-intercept' is 0.48
so log a = 0.48 and a=10^0.48
a= 3.01995172 =3.02

The proposed relation is therefore y = 3.02(2.54)^x
0
6 months ago
#5
(Original post by LukeWatson4590)
Thank you very much for your reply. Sorry to be unclear. The table on the left is the original table of values from the question of the values of variables x and y found in the experiment. You are absolutely right, I thought I had done something silly but could not see why. I have log x instead of log y in the table !
Hopefully correcting this I have attached an amended table and graph.
The gradient I believe can be calculated as approximately:
3.4-0.48/7.2-0
2.92/7.2
0.405555556
Thus, log b =0.405555556 and b = 10^ 0.405555556
b=2.544225233= 2.54

The 'log y-intercept' is 0.48
so log a = 0.48 and a=10^0.48
a= 3.01995172 =3.02

The proposed relation is therefore y = 3.02(2.54)^x
Looks ok. Your method was good last time, apart from the obvious slip.
Just check by evaluating exponential model on the original data, and from a couple of points, it Looks good?
0
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