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Four Logarithmic Statements and Which Three Fail | JEE Advanced Logarithms

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Question
Which of the following statements does not hold good
Alog⁡10(1.42−1)\log_{10}\left(1.4^{2}-1\right) is positivecorrect
Bthe equation log⁡a(a+2)=2\log_{a}(a+2) = 2 is satisfied by two integral values of aacorrect
Clog⁡0.1cot⁡3π8\log_{0.1}\cot\dfrac{3\pi}{8} is negativecorrect
Dif m=4log⁡47m = 4^{\log_{4}7} and n=(19)−2log⁡37n = \left(\dfrac19\right)^{-2\log_{3}7} then n=m4n = m^{4}
Solution
Step 1: Test (1). Compute the argument first.
1.42−1=1.96−1=0.96.1.4^{2}-1 = 1.96-1 = 0.96 .
 Since 0.96<10.96 < 1 and the base 1010 exceeds 11,
log⁡10(0.96)<0,\log_{10}(0.96) < 0 ,
 so the statement "is positive" is false - it does **not** hold good. (1) Step 2: Test (2). Convert the logarithmic equation.
log⁡a(a+2)=2 ⟹ a+2=a2 ⟹ a2−a−2=0 ⟹ (a−2)(a+1)=0.\log_{a}(a+2) = 2 \ \Longrightarrow\ a+2 = a^{2} \ \Longrightarrow\ a^{2}-a-2 = 0 \ \Longrightarrow\ (a-2)(a+1) = 0 .
 Step 3: Apply the base restrictions to those roots. A base must satisfy a>0a>0 and a≠1a \ne 1, so a=−1a = -1 is rejected and only a=2a = 2 has one integral value, not two. The statement does **not** hold good. (2) Step 4: Test (3). Simplify the argument.
cot⁡3π8=cot⁡67.5∘=tan⁡22.5∘=2−1≈0.414.\cot\frac{3\pi}{8} = \cot 67.5^{\circ} = \tan 22.5^{\circ} = \sqrt2 - 1 \approx 0.414 .
 Step 5: Determine the sign. The base 0.10.1 is less than 11, so log⁡0.1\log_{0.1} is **decreasing** and is positive on (0,1)(0,1). Since 0<2−1<10 < \sqrt2-1 < 1,
log⁡0.1(2−1)>0,\log_{0.1}\left(\sqrt2-1\right) > 0 ,
 so calling it negative is false - it does **not** hold good. (3) Step 6: Test (4). Simplify each quantity using alog⁡aN=Na^{\log_{a}N} = N.
m=4log⁡47=7.m = 4^{\log_{4}7} = 7 .
n=(19)−2log⁡37=92log⁡37=(32)2log⁡37=34log⁡37=(3log⁡37)4=74.n = \left(\frac19\right)^{-2\log_{3}7} = 9^{2\log_{3}7} = \left(3^{2}\right)^{2\log_{3}7} = 3^{4\log_{3}7} = \left(3^{\log_{3}7}\right)^{4} = 7^{4} .
 Step 7: Compare. m4=74=nm^{4} = 7^{4} = n, so (4) is a **true** statement and does hold good. Answer: (1), (2) and (3).
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