How do you solve #81^x = 9(3^x)#?

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Answer 1 · 2015-12-04T08:12:09

Apply properties of logarithms and exponents to find
#x = 2/3#

For this problem, we will be using the following:

#81^x = 9(3^x)#

#=> (3^4)^x = 3^2(3^x)#

#=> 3^(4x) = 3^(x+2)#

#=> log_3(3^(4x)) = log_3(3^(x+2))#

#=> 4xlog_3(3) = (x+2)log_3(3)#

#=> 4x(1) = (x+2)(1)#

#=> 4x = x+2#

#=> 3x = 2#

#=> x = 2/3#