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Chris Fox's Engineering Section

Exponential Functions and Logarithms

An exponential function, of the form y=kax, is one in which a given number is repeatedly multiplied by itself.

Consider such a function where a number, denoted by "a", appears "b" times. If this function is multiplied by one in which "a" appears "c" times, in the new function "a" appears b+c times. Therefore, abac=ab+c. This is one of the Laws of Indices.

To find the other, multiply the first function by itself several times so that it appears "c" times. "a" now appears bc times. Therefore, (ab)c=abc.

One property of an exponential curve is that the gradient is directly proportional to the y value (dy/dx ∝ y). There is a constant e, defined such that d(ex)/dx=ex. The value of e is approximately 2.71828, and ex may also be written as ΣN=0(xN/N!), where Σ denotes "sum of" and N! (N factorial) = 1x2x3x...xN.

The inverse of the exponential function is the logarithm, written in the form logab (log to base a of b). If y=ax, then x=logay. The base may be omitted if its value is 10, while loge is termed the natural log, denoted by ln.

The function y=kax can now be transformed into one that is easy to differentiate. Let ax=ebx. This becomes (eb)x. Therefore, a=eb and b=ln a. The original equation can now be written as y=kex ln a, and the gradient is k(ln a)ex ln a.

To differentiate ln x, remember that if y=ex, dy/dx=ex. Therefore, dy/dx=y and dx/dy=y-1. Because y=ex, x=ln y. Therefore, d(ln y)/dy=dx/dy=y-1.

To differentiate y=xx, start by rewriting the right hand side as follows: xx=(eln x)x=ex ln x. Taking logs of both sides gives ln y=x ln x. If both sides are equal to another variable u, then du/dy=y-1 and, using the product rule, du/dx=1+ln x. Finally, using the chain rule and substituting gives dy/dx=(dy/du)(du/dx)=xx(1+ln x).

The hyperbolic functions can be written in terms of e. cosh x=(ex+e-x)/2 and sinh x=(ex-e-x)/2. As with the trigonometric functions, tanh x=(sinh x/cosh x). It will be noticed that cosh x+sinh x=ex and cosh x-sinh x=e-x.

The inverse hyperbolic cosine can be calculated as follows. If y=cosh-1 x, x=cosh y=(ey+e-y)/2. If Y=ey, x=(Y+Y-1)/2. Rearranging gives Y²-2xY+1=0, and solving gives Y=x+√(x²-1). Because y=ln Y, y=cosh-1 x=ln (x+√(x²-1)). Similarly, sinh-1 x=ln (x+√(x²+1)) and tanh-1 x=0.5 ln ((1+x)/(1-x)).

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