Question

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

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

Answer 1

Answer: (A)

$\log b$

$a,b,c$ are consecutive numbers.
Let $a=n-1,b=n,c=n+1$
We have,
$\log (1+ac)=2K$

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

$\Rightarrow \log (1+n^2-1)=2K$

$\Rightarrow \log \left(n^2\right)=2K$
$\Rightarrow 2\log n=2K,(\because \log x^p=p\log x)$
$\Rightarrow K=\log n$
$\Rightarrow K=\log b$

So, option $A$ is correct