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CAT 2021 Slot 1 — QA Question 10

LogarithmsEasy

If 5 – log10(1+x) + 4log10(1-x) = log10(11-x2), then 100x equals

Answer & solution

Answer: 99

Solution

Easy

Rewrite the square roots as half-powers, pull exponents out of the logs, and combine everything over base 1010. The 1+x\sqrt{1+x} terms cancel, leaving a single log of a power of (1−x)(1-x) — then exponentiate.

1

Convert roots to powers and apply log⁡ak=klog⁡a\log a^{k}=k\log a. Write 1+x=(1+x)1/2\sqrt{1+x}=(1+x)^{1/2}, 1−x=(1−x)1/2\sqrt{1-x}=(1-x)^{1/2}, and the right side 11−x2=(1−x2)−1/2\dfrac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}.

5−12log⁡(1+x)+4⋅12log⁡(1−x)=−12log⁡(1−x2)⇒ 5−12log⁡(1+x)+2log⁡(1−x)=−12log⁡(1−x2)\begin{aligned} &5-\tfrac12\log(1+x)+4\cdot\tfrac12\log(1-x)=-\tfrac12\log(1-x^{2})\\ &\Rightarrow\ 5-\tfrac12\log(1+x)+2\log(1-x)=-\tfrac12\log(1-x^{2}) \end{aligned}

(all logs base 1010)

2

Expand the right side. 1−x2=(1+x)(1−x)1-x^{2}=(1+x)(1-x), so −12log⁡(1−x2)=−12log⁡(1+x)−12log⁡(1−x)-\tfrac12\log(1-x^{2})=-\tfrac12\log(1+x)-\tfrac12\log(1-x).

5−12log⁡(1+x)+2log⁡(1−x)=−12log⁡(1+x)−12log⁡(1−x)\begin{aligned} &5-\tfrac12\log(1+x)+2\log(1-x)=-\tfrac12\log(1+x)-\tfrac12\log(1-x) \end{aligned}
3

Cancel log⁡(1+x)\log(1+x) and collect log⁡(1−x)\log(1-x). The −12log⁡(1+x)-\tfrac12\log(1+x) appears on both sides.

5+2log⁡(1−x)=−12log⁡(1−x)⇒ 5=−12log⁡(1−x)−2log⁡(1−x)⇒ 5=−52log⁡(1−x)\begin{aligned} &5+2\log(1-x)=-\tfrac12\log(1-x)\\ &\Rightarrow\ 5=-\tfrac12\log(1-x)-2\log(1-x)\\ &\Rightarrow\ 5=-\tfrac52\log(1-x) \end{aligned}
4

Solve for xx, then 100x100x.

log⁡(1−x)=−2⇒ 1−x=10−2=1100⇒ 100−100x=1⇒ 100x=99\begin{aligned} &\log(1-x)=-2\\ &\Rightarrow\ 1-x=10^{-2}=\tfrac{1}{100}\\ &\Rightarrow\ 100-100x=1\\ &\Rightarrow\ 100x=99 \end{aligned}
100x=99100x=99
CAT 2021 Slot 1 QA Q10: If 5 – log 10 1 + x + 4 log 10 1 - x = log 10 1 1 - x 2 , then 100x equals — Solution | TheCATExam