## 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, $$\log\frac{1}{100} = \log 0.01$$ = $$\log 10^{-2} = -2\log 10 = -2.$$