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CAT 2023 Slot 2 — QA Question 7

LogarithmsEasy

For some positive ral number x, if log3 (x) + logx 25logx (0.008) = 163, then the value of log3 (3x2) is

Answer & solution

Correct answer: 7

Answer: 7

Solution

Easy

Convert every log to base-form you can simplify: log⁡3x=2log⁡3x\log_{\sqrt3}x=2\log_3 x, and the ratio log⁡x25log⁡x0.008\dfrac{\log_x 25}{\log_x 0.008} is just log⁡0.00825\log_{0.008}25. Reduce to find xx, then evaluate log⁡3(3x2)\log_3(3x^2).

1

Simplify the second term (change of base cancels log⁡x\log_x), noting 0.008=81000=5−30.008=\frac{8}{1000}=5^{-3} and 25=5225=5^2:

log⁡x25log⁡x0.008=log⁡0.00825=log⁡52log⁡5−3=2−3=−23\begin{aligned} &\frac{\log_x 25}{\log_x 0.008}=\log_{0.008}25=\frac{\log 5^2}{\log 5^{-3}}=\frac{2}{-3}=-\frac23 \end{aligned}
2

Simplify the first term and substitute into the given equation:

log⁡3x=log⁡31/2x=2log⁡3x⇒ 2log⁡3x−23=163\begin{aligned} &\log_{\sqrt3}x=\log_{3^{1/2}}x=2\log_3 x\\ &\Rightarrow\ 2\log_3 x-\frac23=\frac{16}{3} \end{aligned}
3

Solve for log⁡3x\log_3 x:

2log⁡3x=163+23=183=6⇒ log⁡3x=3 ⇒ x=27\begin{aligned} &2\log_3 x=\frac{16}{3}+\frac{2}{3}=\frac{18}{3}=6\\ &\Rightarrow\ \log_3 x=3\ \Rightarrow\ x=27 \end{aligned}
4

Evaluate the required expression:

log⁡3(3x2)=log⁡3 ⁣(3⋅272)=log⁡3 ⁣(3⋅36)=log⁡337=7\begin{aligned} &\log_3(3x^2)=\log_3\!\left(3\cdot 27^2\right)=\log_3\!\left(3\cdot 3^6\right)=\log_3 3^7=7 \end{aligned}
log⁡3(3x2)=7\log_3(3x^2)=\mathbf{7}

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CAT 2023 Slot 2 QA Q7 — Logarithms solution | TheCATExam