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  • 23-02-2017
  • Mathematics
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3^n+4=27^2n

solve using common bases?

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jdoe0001 jdoe0001
  • 23-02-2017
[tex]\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[/tex]

pretty sure you'd know what "n" is
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