**Log-log graphs. **

We discussed in the last post that a straight line graph is very helpful in interpreting experimental data but sometimes experiments give us data that represent curves on the graph. These curves are not very helpful as they do not give us the exact relationship between our two variables x and y.

x and y can be related in the following two ways –

1) **y=mx+b** ..This is a **linear** relationship and as discussed in the earlier post, this equation gives us a** straight line**.

2)**y=Ax**^{n }**..**This** **is an **exponential** relationship.. This equation gives us a** curve. **

**Point to remember – **

**Point to remember –**

**y=mx+b ****⇒**** straight line ⇒ linear relationship**

y=Ax^{n} ⇒ Curve ⇒ Exponential relationship

If n=2 , equation is a *quadratic* equation i.e y=Ax^{2}

If n=4,equation is a *quartic* equation i.e y=Ax^{4.}

Both these equations give us 2 different curves on the graph as shown in the following fig-

So, if we plot experimental data on a graph paper and we get a curve , we cannot determine the exact relationship between x and y (it could be y=Ax^{2} or y=Ax^{3} or y=Ax^{4} etc). Thus,**curves are not very helpful in interpretation of data** as they tell us nothing accurate.But,straight lines give us exact relation between the two variables under study, x and y. **So, how can we convert the curves to a straight line? The answer is by taking logarithms on both sides!**

y=Ax^{n}

Taking log on both sides, we get ,

log(y) = log(xn) +log A , A=0(y intercept) when line passes through origin.

**log(y) = n log x ** [As log(xn) = n log x] – This is now a linear equation of type **y=mx** !

If we plot a graph of log(y) vs log (x), we would get a straight line passing through origin!

**Note** ⇒ In the above graph, Slope (m) = ‘n’ the exponential factor.

Slope(m) of this straight line can easily be determined by choosing any two points on the line and plugging in values in the equation , m = y2-y1 / x2-x1. Here, the slope is nothing but the exponential factor ’n’.

Thus , if slope is 2 we know y=x^{2 }or if slope is 3 , we know y=x^{3 . }

If, the line does not pass through the origin , i.e it has a y intercept then , the equation shall become ,

**log(y) = n log(x) + log A** , which is like the linear equation, **y=mx + b**. ‘log A’ is the y- intercept and it is a constant.The graph will be as follows –

Thus, **a logarithmic function has an exponential function as its inverse. **

**y= log **_{10}** x implies y=10**^{x}** and **

**ln **_{e}** x implies , y=e**^{x.}

So, for exponential functions we take the logarithm on both sides and convert the equation into a linear one . We further find the slope of the line and determine the value of exponential factor n. After taking log if we do not get a straight line, it means that the two variables are not related exponentially.

**Point to remember** → We take logarithm of an exponential function to get a straight line on the graph.

**Point to remember**

We shall study these kind of log – log graphs later in Chemical Kinetics and Nuclear Chemistry in future posts.Till then,

Be a perpetual student of life and keep learning..

Good Day .