How do you solve $3^{n - 2} = 27$ $3^(n-2)=27$ ?

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Answer 1 · 2016-10-13T10:22:24

$n = 5$ $n=5$

As $3^{n - 2} = 27$ $3^(n-2)=27$

converting it into logarithmic form, we have

${log}_{3} 27 = n - 2$ $log_3 27=n-2$

But ${log}_{3} 27 = {log}_{3} 3^{3} = 3 {log}_{3} 3 = 3 \times 1 = 3$ $log_3 27=log_3 3^3=3log_3 3=3xx1=3$

Hence $n - 2 = 3$ $n-2=3$

and $n = 3 + 2 = 5$ $n=3+2=5$

How do you solve 3n−2=273^(n-2)=27?