ABC26GN2072 · Logarithms

Subject: General Aptitude · Chapter: Logarithms · Exam: 2007 · Marks: · Difficulty:

If $\log \frac{a}{b}+\log \frac{b}{a}=\log (a+b)$, then
(a)$a+b=1$
(b)$a-b=1$
(c)$a=b$
(d)$a^{2}-b^{2}=1$
Answer
Answer (as printed): A
Explanation
$\log \frac{a}{b}+\log \frac{b}{a}=\log (a+b) \Rightarrow \log (a+b)=\log \left(\frac{a}{b} \times \frac{b}{a}\right)=\log 1$. $$\text { So, } a+b=1 \text {. }$$

Explanation as extracted from the printed page; notation may be imperfect.

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