How do you evaluate ${log}_{3} 3 \sqrt[4]{3}$ $log_3 3root4 [3]$ ?

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How do you evaluate ${log}_{3} 3 \sqrt[4]{3}$ $log_3 3root4 [3]$ ?

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Answer 1 · 2016-03-23T18:11:46

I found $\frac{5}{4}$ $5/4$

Explanation:

If I got the idea right I would write it as:
${log}_{3} (3 \cdot 3^{\frac{1}{4}}) = {log}_{3} (3^{1 + \frac{1}{4}}) = {log}_{3} (3^{\frac{5}{4}}) = \frac{5}{4}$ $log_3(3*3^(1/4))=log_3(3^(1+1/4))=log_3(3^(5/4))=5/4$
using the definition of log:
${log}_{a} b = x \to a^{x} = b$ $log_ab=x ->a^x=b$

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How do you evaluate ${log}_{3} 3 \sqrt[4]{3}$ $log_3 3root4 [3]$ ?

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How do you evaluate ${log}_{3} 3 \sqrt[4]{3}$ $log_3 3root4 [3]$ ?