1.7 Raising Numbers to Powers

Starting with our first equation in this chapter, we had \(x=b^p\) and we said that the logarithm of \(x\) is \(\log_b x = p\). Now suppose we take \(x\) and raise it to the power of \(r\), \[ x^r = (b^p)^r = b^{p\times r}, \] and then take the logarithm: \[ \log_b x^r = p\times r. \] Since \(p = \log_b x\), we have the general result: \[ \log_b x^r = r\times \log_b x. \]

To raise a number \(x\) to some arbitrary power \(r\), one would take the logarithm of \(x\), find the product of that number times \(r\), and then find the number whose logarithm is equal to that product.

Suppose we wish to compute \(y = 7^{2/5}\). Finding the logarithm of each side, we get

\[ \log y = \log 7^{2/5} = 2/5 \times \log 7 = 0.4 \times 0.84510 = 0.33804 \]

The number whose log is 0.33804 is then looked up in a table, say, and found to be \(y\) = 2.1779. So, \(7^{2/5}\) = 2.1779.

As a check, we can compute \(y^{5/2} = (\sqrt y)^5\) on a computer, which should give us “7” to good accuracy:

## [1] sqrt(2.1779)*sqrt(2.1779)*sqrt(2.1779)*sqrt(2.1779)*sqrt(2.1779)  =  7.0000

For the case where an exponent is a negative number, our relationship shows that division is implied, as the resulting logarithm will be subtracted in any ongoing calculation. For example,

\[ \log \frac{1}{x^r} = \log x^{-r} = -r\times \log x. \]

This emphasizes that a number less than one will have a negative logarithm. For instance, suppose we have \(u = x^r\), where \(x\) and \(r\) are each greater than one. Then \(u>1\). Now the number \(v =1/u < 1\) and its logarithm will be \(\log v = \log x^{-r}\) = \(-r\cdot\log x < 0.\) Consider, for example, \(\log\frac{1}{100} = \log 0.01\) = \(\log 10^{-2} = -2\log 10 = -2.\)