How do you solve #9^(x+1) = 27#?

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Answer 1 · 2016-04-12T01:48:09

#x = 1/2#

We will use the following properties:

#9^(x+1) = 27#

#=> log_3(9^(x+1)) = log_3(27)#

#=>(x+1)log_3(9) = log_3(27)#

#=>(x+1)log_3(3^2) = log_3(3^3)#

#=>2(x+1) = 3#

#=>x+1 = 3/2#

#:. x = 1/2#

Answer 2 · 2016-04-12T01:51:08

Alternative solution:

You can write everything in the same base:

#(3^2)^(x + 1) = 3^3#

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

#3^(2x) = 3^(3 - 2)#

#3^(2x) = 3^1#

#2x = 1#

#x = 1/2#

Hopefully this helps!