How do you evalute ${log}_{3} (18)$ $Log_ 3 (18)$ ?

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How do you evalute ${log}_{3} (18)$ $Log_ 3 (18)$ ?

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Answer 1 · 2016-01-01T21:18:07

${log}_{3} (18) = 2 + {log}_{3} (2) \approx 2.6309$ $log_3(18)=2+log_3(2)approx2.6309$

Explanation:

${log}_{3} (18) = {log}_{3} (3^{2} \times 2)$ $log_3(18)=log_3(3^2xx2)$

Use the rule: ${log}_{a} (b c) = {log}_{a} (b) + {log}_{a} (c)$ $log_a(bc)=log_a(b)+log_a(c)$

$\Rightarrow {log}_{3} (3^{2}) + {log}_{3} (2)$ $=>log_3(3^2)+log_3(2)$

Use the rule: ${log}_{a} (a^{b}) = b$ $log_a(a^b)=b$

$\Rightarrow 2 + {log}_{3} (2)$ $=>2+log_3(2)$

This is a simplified answer. However, if you want your answer in decimal form, use a calculator. Since many calculators don't have the capability of finding logarithms with a specific base, use the change of base formula.

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

Thus, since the $ln$ $ln$ button is on most calculators (or just the $log$ $log$ button),

${log}_{3} (2) = \frac{ln (2)}{ln (3)} = \frac{log (2)}{log (3)} \approx 0.6309$ $log_3(2)=ln(2)/ln(3)=log(2)/log(3)approx0.6309$

Thus,

${log}_{3} (18) = 2 + {log}_{3} (2) \approx 2.6309$ $log_3(18)=2+log_3(2)approx2.6309$

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