Question

If $\log 2=a$ and $\log 3=b$, then the value of $\left[\log \right(1)+\log (1+3)+\log (1+3+5)+....+....+\log (1+3+5+7+....+19\left)\right]$

$-2[\log 1+\log 2+\log 3+....\log 7]=p+qa+rb$, where $p,q,r,$ are constants. What is the value of $p+2q+3r$ if all logs are in base $10$?

• A. $12$
• B. $26$
• C. $18$
• D. $29.397$

Answer 1

Answer: (D)

$29.397$

We know:

$1=1^2$

$1+3=4=2^2$

$1+3+5=9=3^2$

$1+3+5+7=16=4^2$

....

$1+3+5+7+...+13=49=7^2$

$1+3+5+7+...+13+15=64=8^2$

$1+3+5+7+...+13+15+17=81=9^2$

$1+3+5+7+...+13+15+17+19=100={10}^2$

Replacing the value in the equation, we get:

$\therefore \left[\log \right(1^2)+\log (2^2)+...+\log (8^2)+\log (9^2)+\log ({10}^2\left)\right]-\left[\log \right(1^2)+\log (2^2)+...+\log (7^2\left)\right]=p+qa+rb$

$\Rightarrow \log \left(8^2\right)+\log \left(9^2\right)+\log \left({10}^2\right)=p+qa+rb$

$\Rightarrow \log \left(2^6\right)+\log \left(3^4\right)+2\log \left(2\right)+2\log \left(5\right)=p+qa+rb$

$\Rightarrow 6\log \left(2\right)+4\log \left(3\right)+2\log \left(2\right)+2\log \left(5\right)=p+qa+rb$

$\Rightarrow 2\log \left(5\right)+8a+4b=p+qa+rb$

$\therefore p=2\log \left(5\right),q=8,r=4$

$\therefore p+2q+3r$

$=2\log \left(5\right)+2\times 8+3\times 4$

$=29.397$