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Logarithm question

2023 · 11 Apr · Shift 1 · Q39
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  5. /2023 · 11 Apr · Shift 1 · Q39

Logarithm question

2023 · 11 Apr · Shift 1 · Q39

JEE MainMathematicsLogarithmMCQ+4 / −1
The number of integral solutions xxx of log⁡(x+72)(x−72x−3)2≥0\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0log(x+27​)​(2x−3x−7​)2≥0 is :
  1. A
    8
  2. B
    7
  3. C
    5
  4. D
    6
View written solutionFree

Correct answer: D

Let a=x+72,b=(x−72x−3)2.a=x+\frac{7}{2},\qquad b=\left(\frac{x-7}{2x-3}\right)^2.a=x+27​,b=(2x−3x−7​)2. We need the integer solutions of log⁡ab≥0.\log_a b\ge 0.loga​b≥0.

For a logarithm to be defined, we need:

  1. a>0a>0a>0
  2. a≠1a\ne 1a=1
  3. b>0b>0b>0

Also, depending on the base:

  • if a>1a>1a>1, then log⁡ab≥0  ⟺  b≥1\log_a b\ge 0 \iff b\ge 1loga​b≥0⟺b≥1
  • if 0<a<10<a<10<a<1, then log⁡ab≥0  ⟺  0<b≤1\log_a b\ge 0 \iff 0<b\le 1loga​b≥0⟺0<b≤1

1. Domain conditions

Base conditions

x+72>0  ⟹  x>−72x+\frac{7}{2}>0\implies x>-\frac{7}{2}x+27​>0⟹x>−27​ So for integer xxx, x≥−3.x\ge -3.x≥−3.

Also, x+72≠1  ⟹  x≠−52,x+\frac{7}{2}\ne 1\implies x\ne -\frac{5}{2},x+27​=1⟹x=−25​, which is not an integer, so no extra restriction.

Argument conditions

b=(x−72x−3)2>0b=\left(\frac{x-7}{2x-3}\right)^2>0b=(2x−3x−7​)2>0 This requires x≠7,x≠32.x\ne 7,\qquad x\ne \frac32.x=7,x=23​. Since xxx is integer, only x≠7x\ne 7x=7 matters.

So domain for integer xxx is: x≥−3,x≠7.x\ge -3,\quad x\ne 7.x≥−3,x=7.

2. Split by base value

Since a=x+72,a=x+\frac72,a=x+27​, we check when 0<a<10<a<10<a<1 and when a>1a>1a>1.

For integer x≥−3x\ge -3x≥−3:

  • x=−3⇒a=12x=-3 \Rightarrow a=\frac12x=−3⇒a=21​, so 0<a<10<a<10<a<1
  • x=−2⇒a=32>1x=-2 \Rightarrow a=\frac32>1x=−2⇒a=23​>1
  • all larger integers also give a>1a>1a>1

Thus only x=−3x=-3x=−3 lies in the case 0<a<10<a<10<a<1.

3. Case I: x=−3x=-3x=−3 (base between 0 and 1)

Then

\left(\frac{-10}{-9}\right)^2=\frac{100}{81}>1.$$ For $0<a<1$, we need $0<b\le 1$ for $\log_a b\ge 0$. But here $b>1$, so this does **not** satisfy the inequality. Hence, $$x=-3 \text{ is not a solution.}$$ --- ## 4. Case II: $x\ge -2$, $x\ne 7$ (base $>1$) Now we need $$b\ge 1$$ that is, $$\left(\frac{x-7}{2x-3}\right)^2\ge 1.$$ Since both sides are nonnegative, $$\left|\frac{x-7}{2x-3}\right|\ge 1 \iff |x-7|\ge |2x-3|.$$ Square both sides: $$(x-7)^2\ge (2x-3)^2.$$ Expand: $$x^2-14x+49\ge 4x^2-12x+9$$ $$0\ge 3x^2+2x-40$$ $$3x^2+2x-40\le 0.$$ Factor: $$3x^2+2x-40=(3x-10)(x+4).$$ So $$(3x-10)(x+4)\le 0.$$ Hence, $$-4\le x\le \frac{10}{3}.$$ Since in this case we already have $x\ge -2$, the integer values are $$x=-2,-1,0,1,2,3.$$ Also $x\ne 7$ is already satisfied. So there are $$6$$ integral solutions. --- ## 5. Verify by direct check of these values For reassurance: - $x=-2$: $\left(\frac{-9}{-7}\right)^2>1$ - $x=-1$: $\left(\frac{-8}{-5}\right)^2>1$ - $x=0$: $\left(\frac{-7}{-3}\right)^2>1$ - $x=1$: $\left(\frac{-6}{-1}\right)^2>1$ - $x=2$: $\left(\frac{-5}{1}\right)^2>1$ - $x=3$: $\left(\frac{-4}{3}\right)^2>1$ All valid, and for $x=4$, $$\left(\frac{-3}{5}\right)^2<1,$$ so it fails. Thus the count $6$ is correct. --- ## Final Answer The number of integral solutions is $$\boxed{6}.$$ So the correct option is **D**.
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