# How to prove is two variables or a graph is exponential

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#1
Suppose we have two graphs of y against x. Graph 1 has the equation y = A . b^-x where and b are constants. Graph 2 has equation y = P . x^-1. Both of the graphs do not show a y intercept on them.

How do we tell them apart? They will look the same. We do it by by choosing 3 values of x which are equally spaced apart. By equally spaced I mean difference between each x value is the same. An example would be x = 2, x = 4 and x = 6.

Suppose we choose three x values X1, X2 and X3. We now need to go and read off the corresponding of y, Y1, Y2 and Y3. If the graph is exponential, Y1/Y2 = Y2/Y3.

This is because Y1 = A . b^-X1, Y2 = A . b^-X2 and so on.

Now: Y1/Y2 = (A . b^-X1)/(A . b^-X2) = b^(-X1 + X2) = b^(X2 - X1). The A s cancel and we can use the law that states a^n / a^m = a^(n-m).

Doing the same for Y2/Y3 gives: Y2/Y3 = (A . b^-X2)/ (A . b^-X3) = b ^(-X2 + X3) = b^(X3 - X2).

Now we are saying that b^(X2 - X1) = b^(X3 - X2). This makes sense because X2 - X1 = X3 - X2, as the difference between X1, X2 and X3 is the same (see the second paragraph).

This method will not work for graph 2. For this graph, Y1 = P/X1, Y2 = P/X2 and so on.

Y1/Y2 = X2/X1 and Y2/Y3 = X3/X2. The two expressions would not be equal.

You do need to know this for A Level.
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2 years ago
#2
Suppose we have two graphs of y against x. Graph 1 has the equation y = A . b^-x where and b are constants. Graph 2 has equation y = P . x^-1. Both of the graphs do not show a y intercept on them.

How do we tell them apart? They will look the same. We do it by by choosing 3 values of x which are equally spaced apart. By equally spaced I mean difference between each x value is the same. An example would be x = 2, x = 4 and x = 6.

Suppose we choose three x values X1, X2 and X3. We now need to go and read off the corresponding of y, Y1, Y2 and Y3. If the graph is exponential, Y1/Y2 = Y2/Y3.

This is because Y1 = A . b^-X1, Y2 = A . b^-X2 and so on.

Now: Y1/Y2 = (A . b^-X1)/(A . b^-X2) = b^(-X1 + X2) = b^(X2 - X1). The A s cancel and we can use the law that states a^n / a^m = a^(n-m).

Doing the same for Y2/Y3 gives: Y2/Y3 = (A . b^-X2)/ (A . b^-X3) = b ^(-X2 + X3) = b^(X3 - X2).

Now we are saying that b^(X2 - X1) = b^(X3 - X2). This makes sense because X2 - X1 = X3 - X2, as the difference between X1, X2 and X3 is the same (see the second paragraph).

This method will not work for graph 2. For this graph, Y1 = P/X1, Y2 = P/X2 and so on.

Y1/Y2 = X2/X1 and Y2/Y3 = X3/X2. The two expressions would not be equal.

You do need to know this for A Level.
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