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

1 Answer
Apr 29, 2017

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