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Solving log base (x+1) of (x - 0.5) = log base (x - 0.5) of (x + 1) | JEE Advanced

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

Question
The equation log⁡x+1(x−0.5)=log⁡x−0.5(x+1)\log_{x+1}(x-0.5) = \log_{x-0.5}(x+1) has
Ano real solution
Bno prime solutioncorrect
Can irrational solution
Dno composite solutioncorrect
Solution
Step 1: Notice that the two sides are reciprocals. Writing u=log⁡x+1(x−0.5)u = \log_{x+1}(x-0.5), the change-of-base rule gives
log⁡x−0.5(x+1)=1u,\log_{x-0.5}(x+1) = \frac{1}{u} ,
 so the equation is
u=1u ⟹ u2=1 ⟹ u=±1.u = \frac1u \ \Longrightarrow\ u^{2} = 1 \ \Longrightarrow\ u = \pm1 .
 Step 2: Try u=1u = 1.
x−0.5=x+1 ⟹ −0.5=1,x - 0.5 = x+1 \ \Longrightarrow\ -0.5 = 1 ,
 impossible, so this branch gives nothing. Step 3: Try u=−1u = -1.
x−0.5=(x+1)−1 ⟹ (x−0.5)(x+1)=1.x-0.5 = (x+1)^{-1} \ \Longrightarrow\ (x-0.5)(x+1) = 1 .
 Step 4: Expand and solve.
x2+0.5x−0.5=1 ⟹ 2x2+x−3=0 ⟹ (2x+3)(x−1)=0,x^{2}+0.5x-0.5 = 1 \ \Longrightarrow\ 2x^{2}+x-3 = 0 \ \Longrightarrow\ (2x+3)(x-1) = 0 ,
x=1orx=−32.x = 1 \quad\text{or}\quad x = -\frac32 .
 Step 5: Check both against the requirements on a logarithm's base - it must be positive and not 11. - x=−32x = -\dfrac32: then x+1=−12<0x+1 = -\dfrac12 < 0, not a valid base. **Reject.** - x=1x = 1: then x+1=2x+1 = 2 and x−0.5=0.5x-0.5 = 0.5, both positive and neither equal to 11. **Accept.** Step 6: Verify.
log⁡2(0.5)=−1,log⁡0.5(2)=−1.\log_{2}(0.5) = -1, \qquad \log_{0.5}(2) = -1 .
 Step 7: Classify the single solution x=1x = 1. - It is real, so (1) is false. - 11 is **not prime**, so the equation has no prime solution. ✓\checkmark (2) - 11 is rational, so (3) is false. - 11 is **not composite**, so the equation has no composite solution. (4) Answer: (2) and (4).
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