Question

If ${\log }_2\left(1+\frac{1}{a}\right)+{\log }_2\left(1+\frac{1}{b}\right)+{\log }_2\left(1+\frac{1}{c}\right)=2$ , where a, b, c are three different positive integers, then ${\log }_e\left(a+b+c\right)={\log }_{e}a+{\log }_{e}b+{\log }_{e}c$ or prove that a = 1, b = 2, c = 3.

• A. True
• B. False

Answer 1

The correct option is A True

Given,

${\log }_2\frac{a+1}{a}+{\log }_2\frac{b+1}{b}+{\log }_2\frac{c+1}{c}=2$

$\Rightarrow {\log }_2\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)=2$

$\Rightarrow \frac{a+1}{a}\times \frac{b+1}{b}\times \frac{c+1}{c}=2^2=4$.........(1)

Since, given equation has $3$ variables but only $1$ equation, so:

Case-$1$: $a+1=b$ and $b+1=c$......................(2)

Hence,$\frac{c+1}{a}=\frac{4}{1}$......................[from $1$]

$\Rightarrow a=1$ and $c=3$.

So, $b=2$..................[from $2$]