How do you solve #3^(x^2 + 20) = (1/27)^(3x)#?

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Answer 1 · 2015-08-20T13:23:14

The solutions are
#color(blue)(x=-4,x=-5#

#3^(x^2+20)=(1/27)^(3x)#

We know that #1/27=1/(3^3#

So,

#3^(x^2+20)=(1/3^3)^(3x)#

By property
#color(blue)(1/a=a^-1#

#3^(x^2+20)=color(blue)((3^-3))^(3x)#

#3^(x^2+20)=3^(-9x)#
Now as bases are equal we equate powers and find #x#

#x^2+20=-9x#

#x^2+9x+20=0#

We can Split the Middle Term of this expression to factorise it and find solutions.

#x^2+color(blue)(9x)+20=0#

#x^2+color(blue)(5x+4x)+20=0#

#x(x+5)+4(x+5)=0#

#(x+4)(x+5)=0#

We now equate the factors to zero.
#x+4=0, x=-4#

#x+5=0, x=-5#