Basics & LogarithmsmediumJEE Main 2020Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2020)

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

Question

The value of

(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:

\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)}

Step 3: Write

0.16=\dfrac{4}{25}

and

2.5=\dfrac{5}{2}

(0.16)^{\log_{5/2}(1/2)}=\left(\dfrac{4}{25}\right)^{\log_{5/2}(1/2)}

Step 4: Note that

\dfrac{4}{25}=\left(\dfrac{2}{5}\right)^{2}=\left(\dfrac{5}{2}\right)^{-2}

. Let

b=\dfrac{5}{2}

=\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

4

. Answer:

4

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