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2x3x4x=3x4x2x2\sqrt[4x]{x^{3x}} = 3\sqrt[2x]{x^{4x}}24xx3x=32xx4x
Find the real value of xxx satisfying the equation above.
The answer is of the form ab5\sqrt[5]{\dfrac{a}{b}}5ba where aaa and bbb are positive co-prime integers, then submit the value of a+ba + ba+b.
Note\text{Note}Note:- Here x≠−1,0,1x \neq -1,0,1x=−1,0,1
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