How do you evaluate $log 900 - log 9$ $log 900 - log 9$ ?

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Answer 1 · 2016-11-06T01:57:46

This can be simplified to $2$ $2$ .

By the rule ${log}_{a} (n) - {log}_{a} (m) = {log}_{a} (\frac{n}{m})$ $log_a(n) - log_a(m) = log_a(n/m)$ , we have:

$log 900 - log 9 = log (\frac{900}{9}) = log 100$ $log900 - log9 = log(900/9) = log100$

Since the notation $log a$ $loga$ is a logarithm in base $10$ $10$ , we can use the change of base formula to rewrite the following:

$= \frac{log 100}{log 10} = \frac{{log 10}^{2}}{log 10} = \frac{2 log 10}{log 10} = 2$ $=log100/log10 = log10^2/log10 = (2log10)/log10 = 2$

Hopefully this helps!

How do you evaluate log900−log9log 900 - log 9?