How do you solve #81^x = 27^(x + 2)#?

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Answer 1 · 2015-07-22T19:25:42

More simply, #x=6#

(logarithms are unnecessary in this case).

#81=3^{4}# and #27=3^{3}#, so this equation can also be written as

#3^{4x}=3^{3(x+2)}=3^{3x+6}#.

Since the bases are now the same, we can equate exponents to get

#4x=3x+6# so that

#x=6#

(technically this equating of exponents requires the fact that exponential functions (with base not equal to 1) are one-to-one functions).