3 years ago

3^n+4=27^2n solve using common bases?

1 answer:

7 0

$\bf 3^{n+4}=27^{2n}\qquad 27=3^3\qquad thus
\\\\
3^{n+4}=(3^3)^{2n}\implies 3^{n+4}=3^{3\cdot 2n}\implies 3^{n+4}=3^{6n}
\\\\
\textit{same bases, thus the exponent must be the same}
\\\\
n+4=6n$

pretty sure you'd know what "n" is

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