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Basics & Logarithms: Sum Possible Values Equation

JEE Maths question with a full step-by-step solution.

Question
Sum of all possible values of xx of the equation (3+2)x+(3−2)x−23=0(\sqrt{3}+\sqrt{2})^x + (\sqrt{3}-\sqrt{2})^x - 2\sqrt{3} = 0 is:
A22
B00correct
C−1-1
D12\dfrac{1}{2}
Solution
Step 1: Use the reciprocal relationship Since (3+2)(3−2)=1(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2}) = 1, let t=(3+2)xt = (\sqrt{3}+\sqrt{2})^x. Then (3−2)x=1t(\sqrt{3}-\sqrt{2})^x = \dfrac{1}{t}. Step 2: Form a quadratic in tt
t+1t=23  ⟹  t2−23 t+1=0t + \frac{1}{t} = 2\sqrt{3} \implies t^2 - 2\sqrt{3}\,t + 1 = 0
t=23±12−42=3±2t = \frac{2\sqrt{3} \pm \sqrt{12-4}}{2} = \sqrt{3} \pm \sqrt{2}
 Step 3: Solve for xx
(3+2)x=3+2  ⟹  x=1(\sqrt{3}+\sqrt{2})^x = \sqrt{3}+\sqrt{2} \implies x = 1
(3+2)x=3−2=(3+2)−1  ⟹  x=−1(\sqrt{3}+\sqrt{2})^x = \sqrt{3}-\sqrt{2} = (\sqrt{3}+\sqrt{2})^{-1} \implies x = -1
 Step 4: Sum the solutions
1+(−1)=01 + (-1) = 0
 Answer: (2)
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