Basics & LogarithmsmediumPYQ · JEE Main · 3 Aug 2020 · Shift 1 (Morning)Free

Basics & Logarithms: Value Equal (JEE Main 2020)

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

Question
The value of (0.16)log⁡2.5 ⁣(13+132+133+⋯ to ∞)(0.16)^{\log_{2.5}\!\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\cdots\text{ to }\infty\right)} is equal to
Solution
Answer: 4
Step 1: Compute the infinite sum inside the logarithm:
13+132+133+⋯=1/31−1/3=12\dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}+\cdots=\dfrac{1/3}{1-1/3}=\dfrac{1}{2}
 Step 2: So the expression becomes:
(0.16)log⁡2.5(1/2)(0.16)^{\log_{2.5}(1/2)}
 Step 3: Write 0.16=4250.16=\dfrac{4}{25} and 2.5=522.5=\dfrac{5}{2}:
(0.16)log⁡5/2(1/2)=(425)log⁡5/2(1/2)(0.16)^{\log_{5/2}(1/2)}=\left(\dfrac{4}{25}\right)^{\log_{5/2}(1/2)}
 Step 4: Note that 425=(25)2=(52)−2\dfrac{4}{25}=\left(\dfrac{2}{5}\right)^{2}=\left(\dfrac{5}{2}\right)^{-2}. Let b=52b=\dfrac{5}{2}:
=(b−2)log⁡b(1/2)=b−2log⁡b(1/2)=blog⁡b(1/2)−2=blog⁡b4=4=\left(b^{-2}\right)^{\log_{b}(1/2)}=b^{-2\log_{b}(1/2)}=b^{\log_{b}(1/2)^{-2}}=b^{\log_{b}4}=4
 Step 5: Therefore the value is 44. Answer: 44
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