How do you find a value of n that makes $4^{3 n} \cdot 4^{n - 2} = 4^{18}$ $4^(3n)*4^(n-2)=4^18$ a true statement?

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Answer 1 · 2016-11-04T01:46:00

$n = 5$ $n=5$

We have:

$4^{3 n} \times 4^{n - 2} = 4^{18}$ $4^(3n)xx4^(n-2)=4^18$

Let's first combine the left side:

$4^{3 n + (n - 2)} = 4^{18}$ $4^(3n+(n-2))=4^18$

We can now take the ${log}_{4}$ $log_4$ of both sides:

${log}_{4} 4^{3 n + (n - 2)} = {log}_{4} 4^{18}$ $log_(4)4^(3n+(n-2))=log_(4)4^18$

$3 n + (n - 2) = 18$ $3n+(n-2)=18$

$4 n - 2 = 18$ $4n-2=18$

$4 n = 20$ $4n=20$

$n = 5$ $n=5$

How do you find a value of n that makes 43n⋅4n−2=4184^(3n)*4^(n-2)=4^18 a true statement?