Often in experiments, when 2 variables are related, a set of corresponding values is obtained. This set of values can be used to draw a graph. If all the points lie on a straight line, the relationship connecting the 2 variables can be known. Such a relationship will be Y = m X + c, where Y and X are the 2 variables and m is the gradient and c the intercept on the Y-axis. But many experiments do not result in a linear relationship. Examples are y = ax² , y = a/x + b, y = ax^b (x^b means x to power of b), etc.

In the case of y = ax², if you plot y against x², a straight line is obtained with gradient a and passing through the origin.

In the case of y = a/x + b, if you plot y against 1/x, a straight line is obtained with gradient a and b is the intercept on the y-axis.

In the case of y = ax^b, we take common logarithms of both side first.

lg(y) = lg(ax^b) ⇒ lg(y) = b.lg(x) + lg(a). Now a straight line graph is obtained, when lg(y) is plotted against lg(x). b will be the gradient and lg(a) is the intecept on the lg(y) – axis.

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*Now, try these questions* :

1) Variables x and y are connected by an equation of the form y = ab^(−x), where a and b are constants. When lg(y) is plotted against x, a straight line is obtained which passes through (2, 1) and (5, −5). Find the values of a and b.

2) The variables x and y are related in such a way that when y/x² is plotted against x, a straight line is obtained which passes through (2, 1) and (7, 6). Express y in terms of x.

*Answers :*

1) Gradient of straight line = (1 + 5)/(2 − 5) = 6/(−3) = −2

Take common log of y = ab^(−x) ⇒ lg(y) = −lg(b).x + lg(a) (note lg(b^(−x)) = −x.lg(b) )

∴ −lg(b) = −2 ⇒ lg(b) = 2 ⇒ **b = 10² = 100**

Since the straight line passes through (2,1), substituting (2,1) into the ‘lg’ equation,

1 = −2 (2) + lg(a) ⇒ lg(a) = 5 ⇒ ** a = 10^5 = 100,000**

2) The straight line will have the equation, y/x² = mx + c

m = (6 − 1)/(7 − 2) = 5/5 = 1

(1 − c)/(2 −0) = 1 ⇒ 1 − c = 2 ⇒ c = −1

∴ y/x² = x − 1 ⇒ **y = x³ − x²**

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