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CAT 2017 Slot 2 — QA Question 26

MeansEasy

If log (2a × 3b × 5c) is the arithmetic mean of log (22 × 33 × 5), log (26 × 3 × 57), and log (2 × 32 × 54), then a equals

Answer & solution

Correct answer: 3

Answer: 3

Solution

Easy

The arithmetic mean of three logs equals one-third the log of the product. Add up the prime exponents, divide by 33, then match the exponent of 22 on both sides.

1

Write the mean as a single logarithm. Use log⁡X+log⁡Y+log⁡Z=log⁡(XYZ)\log X+\log Y+\log Z=\log(XYZ).

AM=13[log⁡(22335)+log⁡(263 57)+log⁡(2 3254)]⇒ AM=13log⁡ ⁣(22335⋅263 57⋅2 3254)\begin{aligned} &\text{AM}=\tfrac13\Big[\log(2^2 3^3 5)+\log(2^6 3\,5^7)+\log(2\,3^2 5^4)\Big]\\ &\Rightarrow\ \text{AM}=\tfrac13\log\!\big(2^2 3^3 5\cdot 2^6 3\,5^7\cdot 2\,3^2 5^4\big) \end{aligned}
2

Add the exponents of each prime. Collect powers of 22, 33 and 55 inside the product.

AM=13log⁡ ⁣(2 2+6+1⋅3 3+1+2⋅5 1+7+4)⇒ AM=13log⁡ ⁣(29⋅36⋅512)⇒ AM=log⁡ ⁣(23⋅32⋅54)(divide each exponent by 3)\begin{aligned} &\text{AM}=\tfrac13\log\!\big(2^{\,2+6+1}\cdot 3^{\,3+1+2}\cdot 5^{\,1+7+4}\big)\\ &\Rightarrow\ \text{AM}=\tfrac13\log\!\big(2^{9}\cdot 3^{6}\cdot 5^{12}\big)\\ &\Rightarrow\ \text{AM}=\log\!\big(2^{3}\cdot 3^{2}\cdot 5^{4}\big) \quad\text{(divide each exponent by 3)} \end{aligned}
3

Match the powers. The mean equals log⁡(2a3b5c)\log(2^a 3^b 5^c), so (taking a,b,ca,b,c as integers) compare exponents.

2a3b5c=233254⇒ a=3,b=2,c=4\begin{aligned} &2^a 3^b 5^c=2^3 3^2 5^4\\ &\Rightarrow\ a=3,\quad b=2,\quad c=4 \end{aligned}
a=3a=3

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CAT 2017 Slot 2 QA Q26: If log (2 a × 3 b × 5 c ) is the arithmetic mean of log (2 2 × 3 3 × 5), log (2 6 &tim — Solution | TheCATExam