How do you solve #27^(x-3)=1/9#?

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Answer 1 · 2015-11-29T09:41:16

Express as powers of #3# and hence find #x = 7/3#

If #a > 0# then #a^(bc) = (a^b)^c#

So:

#3^(3(x-3)) = (3^3)^(x-3) = 27^(x-3) = 1/9 = 3^-2#

Since #f(x) = 3^x# is a one-one function from #RR->(0, oo)#, the Real solutions of this must have equal exponents:

#3(x-3) = -2#

Hence:

#3x = -2+9 = 7#

So #x = 7/3#