SOLUTION: I can not for the life of me remember how to do this. I have tried numerous ways and still have not come to the correct answer. 2x to the 4-3x to the 3+x+1 is divided by 2x to t

Question 33135This question is from textbook college algebra
: I can not for the life of me remember how to do this. I have tried numerous ways and still have not come to the correct answer.
2x to the 4-3x to the 3+x+1 is divided by 2x to the 2+x+1 This question is from textbook college algebra

Answer by mukhopadhyay(490) (Show Source): You can put this solution on YOUR website!
This math looks hard, but easy to solve.
[(2x)^(4-3x)]^(3+x+1) / [(2x)^(2+x+1)];
Based on the property (x^m)^(n) = x^(mn);
So,
[(2x)^(4-3x)]^(3+x+1) = (2x)^[(4-3x)(4+x)]
=> [(2x)^(4-3x)]^(3+x+1) = (2x)^(16+4x-12x-3x^2)
=> [(2x)^(4-3x)]^(3+x+1) = (2x)^(-3x^2-8x+16);
.................
Based on the property: x^m/x^n = x^(m-n);
We have the same base in the question and the common base is 2x;
Thus,
[(2x)^(4-3x)]^(3+x+1) / [(2x)^(2+x+1)] = [(2x)^(-3x^2-8x+16)] / [(2x)^(2+x+1)]
=> [(2x)^(4-3x)]^(3+x+1) / [(2x)^(2+x+1)] = (2x)^[(-3x^2-8x+16) - (3+x)]
=> [(2x)^(4-3x)]^(3+x+1) / [(2x)^(2+x+1)] = (2x)^[(-3x^2-9x+13)]
The answer is (2x)^[(-3x^2-9x+13)]

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