How do you solve #2^ { 2x } - ( 2^ { x + 2} ) - 32= 0#?

1 Answer
Apr 29, 2017

Answer:

#x=3#

Explanation:

First of all, you should know that any number of the form #a^(bc)# can be written in the form #(a^b)^c# or #(a^c)^b# or vice-versa.

i.e. #a^(bc) = (a^b)^c = (a^c)^b#.

Also, #a^(b+c) = a^b*a^c#

#therefore # above equation #2^(2x)-(2^(x+2))-32=0# can be writte as:-

#(2^x)^2 - 2^2*2^x-32=0#

#=> (2^x)^2-4*2^x-32=0#

Let #2^x# be some number #y#.

#therefore y^2-4y-32=0#

Now solving it just like a regular quadratic equation

#=> y^2-8y+4y-32=0#

#=> y(y-8)+4(y-8)=0#

#=> (y+4)(y-8)=0#

#y=-4, 8#

But, #y = 2^x#

#=> 2^x = -4, 8#

Since any exponential function defined on real numbers cannot have a negative value, we will discard #-4#.

graph{2^x [-10, 10, -5, 5]}

You can observe in this graph of #2^x# that no matter what the value of #x#, the value of #y=2^x# is always positive. Its value is getting closer and closer to zero but is never actually touching zero. This is true for any function of the type #a^x# where #a# and #x# are REAL Numbers.

#therefore# #2^x = 8#

#=> 2^x = 2^3#

#=> x=3#