How do you find the exact value of ${log}_{4} 16^{1.2}$ $log_4 16^1.2$ ?

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Answer 1 · 2017-01-18T07:56:44

I got $2.4$ $2.4$

I would first use a property of logs to deal with the exponent of the argument and write:
$1.2 {log}_{4} (16) =$ $1.2log_4(16)=$

Then I would use the definition to solve the log as:
${log}_{4} (16) = 2$ $log_4(16)=2$
[Because $4^{2} = 16$ $4^2=16$ ]

So I have:
$1.2 \cdot 2 = 2.4$ $1.2*2=2.4$

How do you find the exact value of log4161.2log_4 16^1.2?