How do you solve the exponential inequality #27^(4x-1)>=9^(3x+8)#?

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Answer 1 · 2017-03-26T21:03:01

#x>=9.5/3#

#27^(4x-1)>=9^(3x+8)#

law of indices:
#(a^m*a^n=a^(m+n))#

#27 = 3^3#
#9=3^2#

#therefore 27 = 9^(3/2)#

#27^(4x-1) = 9^(1.5(4x-1))#

#=9^(6x-1.5)#

#9^(6x-1.5)>=9^(3x+8)#

#6x-1.5>=3x+8#

add #1.5#:

#6x>=x+9.5#

subtract #3x#:

#3x>=9.5#

#x>=9.5/3#

Answer 2 · 2018-08-10T02:55:26

#x=19/6#

It would be very nice if we had the same base. Recall that

#27=3^3# and #9=3^2#

With this in mind, we can rewrite our inequality as

#3^(3(4x-1))=3^(2(3x+8)#

We can further simplify the exponents to get

#3^(12x-3)=3^(6x+16)#

Now that our bases are the same, we can equate the exponents:

#12x-3=6x+16#

#6x=19#

#x=19/6#

Hope this helps!