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27^{5 x} = 9^{x+3}. Find X
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\(27^{5x} = 9^{x+3}\) It seems like you should make the bases the same, then set the exponents equal to each other and solve for x.
any idea how i could make the bases the same. i tried but failed to do so.
\(3^3=27\) and \(3^2=9\)
Thanks alot
You're Welcome. One last thought... \((3^3)^{5x}=3^{15x}\) When raising a power to a power, you multiply.
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Got it. Thank you again for your help
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