How do you solve #4^x * 2^(x^2) = 16^2#?

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Answer 1 · 2016-05-15T15:42:25

#x=-4# and #x=2#

First, note that the bases of the exponential functions here, which are #4,2,16,# can all be expressed as powers of #2#: #4=2^2# and #16=2^4#.

Rewrite the equation to reflect this:

#(2^2)^x*2^(x^2)=(2^4)^2#

Simplify #(2^2)^x# and #(2^4)^2# using the rule: #(a^b)^c=a^(bc)#.

#2^(2x)*2^(x^2)=2^8#

On the left hand side, simplify the exponents using the rule: #a^b*a^c=a^(b+c)#.

#2^(2x+x^2)=2^8#

Since both of the bases are the same, we know their exponents must also be equal:

#2x+x^2=8#

Solve as you typically would a quadratic equation:

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

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

#x=-4,x=2#