How do you solve #[(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0#?

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Answer 1 · 2018-04-11T14:36:39

#x ~~ .7925#

We can start by dividing one term from both side of the equation and simplifiying:

#[(2^(2x)-3)]*[(2^(x+2))+32] = 0#

#[(2^(2x)-3)]*[(2^(x+2))+32]/[(2^(x+2))+32] = 0/[(2^(x+2))+32]#

#2^(2x)-3 = 0#

#2^(2x) = 3#

Now we can take the Log of the equation with the base being that of the number with the variable in the exponent:

#log_2 2^(2x)=log_2 3#

As you may remember, this is equivalent to:

#2x*(1) = 1.585...#

Finally, solve for #x#

#x ~~ .7925#