Question

If $a,b,c$ are consecutive positive integers and $\log (1+ac)=2k$, then the value of $k$ is

• A. $\log a$
• B. $\log b$
• C. $2$
• D. $1$

Answer 1

The correct option is B

$\log b$

Explanation for the correct option:

Find the value of :

Given $\log (1+ac)=2k$ and $a,b,c$ are consecutive positive integers

Let, $a=n-1,b=n\&c=n+1,wheren>1.$

Then, $\log (1+ac)=2k$ can be written as,

$\log (1+(n-1\left)\right(n+1\left)\right)=2k$

$\Rightarrow$ $\log (1+n^2–1)=2k$

$\Rightarrow$ $\log n^2=2k$

$\Rightarrow$ $2\log n=2k$

$\Rightarrow$ $\log n=k$

$\Rightarrow$ $\mathit{l}\mathit{o}\mathit{g}\mathit{b}\mathbf{=}\mathit{k}$

Hence, correct option is $\left(B\right)$