How to solve #4^(x^2)=(1/32)^((x+6)/5)# ?

1 Answer
Jun 26, 2015

There are no Real solutions to the given equation

Explanation:

#4 = 2^2# and #1/32 = 2^(-5)#

So
#color(white)("XXXX")##4^(x^2) = (1/32)^((x+6)/5)#
is equivalent to
#color(white)("XXXX")##(2^2)^(x^2) = (2^(-5))^((x+6)/5)#
or
#color(white)("XXXX")##2^(2x^2) = 2^(-(x+6))#

So
#color(white)("XXXX")##2x^2 = -(x+6)#

#color(white)("XXXX")##2x^2+x+6 = 0#

Since the discriminant (#b^2-4ac#) is less than #0# there are no Real solutions for the given equation.

Graph of #4^(x^2) - (1/32)^((x+6)/5)# Note that it never reaches the x-axis (i.e. it is never zero).
graph{4^(x^2)-(1/32)^((x+6)/5) [-10, 10, -5, 5]}