How do you solve #(\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}#?

Question

1077 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-20T10:27:03

#x=-4# or #x=-2#

Start by working on the left side:

Since #81=9^2# we know #1/81=1/9^2# and by the rules #1/a=a^-1#, we know that #1/9^2 = 9^-2#

By the rule #(a^b)^c=a^(bc)#, we know

#(9^-2)^(6x+2) =9^(-12x-4)#.

So our equation now looks like:

#9^(-12x-4)=9^(2x^2+12)#

If #a^b = a^c#, then #b=c#, so:

#-12x-4 = 2x^2+12#

moving everything to one side and collecting like terms we have:

#2x^2+12x+16=0#

Dividing thruogh by 2:

#x^2+6x+8=0#

factoring:

#(x+4)(x+2)=0#

Solving:

#x=-4# or #x=-2#