How do you solve #3^(5x)*81^(1-x)=9^(x-3)#?

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Answer 1 · 2016-10-24T02:18:09

Put in a common base.

#3^(5x) xx (3^4)^(1 - x) = (3^2)^(x - 3)#

Simplify using the rules #a^n xx a^m = a^(n + m)# and #(a^n)^m = a^(n xx m)#.

#3^(5x) xx 3^(4 - 4x) = 3^(2x - 6)#

#3^(5x + 4 - 4x) = 3^(2x - 6)#

We can now eliminate the bases and solve like a linear equation.

#5x + 4 - 4x = 2x - 6#

#5x - 4x - 2x = -6 - 4#

#-x = -10#

#x = 10#

Hopefully this helps!

Answer 2 · 2016-10-24T02:23:37

#x = 10#

#3^(5x)*81^(1-x) = 9^(x-3)#

#=> 3^(5x)*(3^4)^(1 - x) = (3^2)^(x-3)#

#=> 3^(5x)*3^(4(1-x)) = 3^(2(x-3))#

#=>3^(5x + 4 - 4x) = 3^(2x - 6)#

#=> 3^(x + 4) = 3^(2x - 6)#

Apply #log_3# to both sides

#=> log_3 3^(x + 4) = log_3 3^(2x - 6)#

#=> (x + 4) log_3 3 = (2x - 6)log_3 3#

#=> x + 4 = 2x - 6#

#=> x = 10#