How do you calculate ${log}_{3} 45 - {log}_{3} 9$ $log_3 45 - log_3 9$ ?

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Answer 1 · 2018-06-26T15:19:05

We will use logarithmic rules to find that the answer is $\frac{log 5}{log 3}$ $log5/log3$ , or about $1.645$ $1.645$ .

The first rule we'll use is ${log}_{a} b - {log}_{a} c = {log}_{a} (\frac{b}{c})$ $log_a b-log_a c=log_a (b/c)$ :

${log}_{3} 45 - {log}_{3} 9 = {log}_{3} (\frac{45}{9}) = {log}_{3} 5$ $log_3 45-log_3 9=log_3 (45/9)=log_3 5$

This logarithm doesn't look too friendly. Fortunately, we can use another rule:

${log}_{b} c = \frac{{log}_{a} c}{{log}_{a} b}$ $log_b c=(log_a c)/(log_a b)$

It doesn't matter what base we choose for $a$ $a$ , as long as each logarithm as the same base:

${log}_{3} 5 = \frac{{log}_{10} 5}{{log}_{10} 3} \approx 1.465$ $log_3 5=(log_10 5)/(log_10 3)~~1.465$

Hope this helps!

How do you calculate log345−log39log_3 45 - log_3 9?