How do you evaluate ${log}_{\sqrt{3}} 243$ $Log_sqrt3 243$ ?

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Answer 1 · 2016-03-13T13:56:53

$10$ $10$

We should try to write $243$ $243$ as a power of $3$ $3$ .

An example of the method we should attempt can be shown more easily in evaluating ${log}_{2} 8$ $log_2 8$ :

${log}_{2} 8 = {log}_{2} 2^{3} = 3 {log}_{2} 2 = 3$ $log_2 8=log_2 2^3=3log_2 2=3$

The first step we should take for ${log}_{\sqrt{3}} 243$ $log_sqrt3 243$ is to recognize that $243 = 3^{5}$ $243=3^5$ .

${log}_{\sqrt{3}} 243 = {log}_{\sqrt{3}} 3^{5}$ $log_sqrt3 243=log_sqrt3 3^5$

But we still haven't addressed the issue that we want a base of $\sqrt{3}$ $sqrt3$ , not $3$ $3$ .

We should use the fact that ${(\sqrt{3})}^{2} = 3$ $(sqrt3)^2=3$ , like so:

${log}_{\sqrt{3}} 3^{5} = {log}_{\sqrt{3}} {({(\sqrt{3})}^{2})}^{5}$ $log_sqrt3 3^5=log_sqrt3 ((sqrt3)^2)^5$

Using the rule that ${(x^{a})}^{b} = x^{a b}$ $(x^a)^b=x^(ab)$ , we multiply $2$ $2$ and $5$ $5$ to see that

${log}_{\sqrt{3}} {({(\sqrt{3})}^{2})}^{5} = {log}_{\sqrt{3}} {(\sqrt{3})}^{10}$ $log_sqrt3 ((sqrt3)^2)^5=log_sqrt3 (sqrt3)^10$

Now, simplify as was done earlier.

${log}_{\sqrt{3}} {(\sqrt{3})}^{10} = 10 {log}_{\sqrt{3}} \sqrt{3} = 10$ $log_sqrt3 (sqrt3)^10=10log_sqrt3sqrt3=10$

How do you evaluate Log_sqrt3 243 log√3243Log_sqrt3 243 ?