Basics & LogarithmseasyFree

Basics & Logarithms: Question

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

Question
If log⁡72x−62x−2>0\log_7\dfrac{2x-6}{2x-2} > 0, then x∈x \in:
A(−∞,1)(-\infty, 1)correct
B[1,2][1, 2]
C(2,∞)(2, \infty)
D[0,2][0, 2]
Solution
Step 1: Remove the logarithm Since the base 7>17 > 1:
2x−62x−2>70=1\frac{2x-6}{2x-2} > 7^0 = 1
 Step 2: Solve
2x−62x−2−1>0=−42x−2>0=−2x−1>0  ⟹  x<1\frac{2x-6}{2x-2} - 1 > 0 = \frac{-4}{2x-2} > 0 = \frac{-2}{x-1} > 0 \implies x < 1
x∈(−∞,1)x \in (-\infty, 1)
 Answer: (1)
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