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Question 80
If $$(\sqrt{2})^{x + 5} = (\sqrt[4]{2})^{2x^2 - 2}$$, then a value of $$(x^2 - 1)$$ is
Solution
The RHS of the equation can be re written as $$\left(\sqrt{\ 2}\right)^{\frac{2x^2-2}{2}}=\left(\sqrt{\ 2}\right)^{x^2-1}$$
As the bases in both LHS and RHS are now same, the powers can be equated.
$$x+5=x^2-1$$
solving this equation gives us x = 3 or x = -2
For the desired value, we get either 8 or 3. Option having 8 is present, and hence, the answer.
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