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CAT 2024 Slot 1 — QA Question 22

LogarithmsEasy

If x is a positive number such that 4 log10x + 4 log100x + 8 log1000x = 13, then the greatest integer not exceeding x, is

Answer & solution

Correct answer: 31

Answer: 31

Solution

Medium

All three logs share the argument xx but have bases that are powers of 10. Convert every term to log⁡10x\log_{10}x, solve the linear equation, then evaluate xx and floor it.

1

Reduce to one base using log⁡10kx=1klog⁡10x\log_{10^k}x=\tfrac{1}{k}\log_{10}x:

L=log⁡10x⇒ log⁡100x=L2(base 102)⇒ log⁡1000x=L3(base 103)\begin{aligned} &L=\log_{10}x\\ &\Rightarrow\ \log_{100}x=\tfrac{L}{2} \quad\text{(base } 10^2\text{)}\\ &\Rightarrow\ \log_{1000}x=\tfrac{L}{3} \quad\text{(base } 10^3\text{)} \end{aligned}
2

Substitute step 1 into the given equation:

4L+4 ⁣(L2)+8 ⁣(L3)=13⇒ 4L+2L+8L3=13\begin{aligned} &4L+4\!\left(\tfrac{L}{2}\right)+8\!\left(\tfrac{L}{3}\right)=13\\ &\Rightarrow\ 4L+2L+\tfrac{8L}{3}=13 \end{aligned}
3

Solve for LL:

6L+8L3=26L3=13⇒ L=32\begin{aligned} &6L+\tfrac{8L}{3}=\tfrac{26L}{3}=13\\ &\Rightarrow\ L=\tfrac{3}{2} \end{aligned}
4

Evaluate and floor using L=32L=\tfrac32:

x=10L=103/2=1010≈31.62⇒ ⌊x⌋=31\begin{aligned} &x=10^{L}=10^{3/2}=10\sqrt{10}\approx31.62\\ &\Rightarrow\ \lfloor x\rfloor=31 \end{aligned}
⌊x⌋=31\lfloor x\rfloor=\mathbf{31}

log⁡10kx=1klog⁡10x\log_{10^k}x=\tfrac1k\log_{10}x collapses any “same number, base a power of 10” sum into one multiple of log⁡10x\log_{10}x; here the coefficient is 4+2+83=2634+2+\tfrac83=\tfrac{26}{3}.

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CAT 2024 Slot 1 QA Q22 — Logarithms solution | TheCATExam