Question

If $\log 2=aand\log 3=b$ then $[\log \left(1\right)+\log (1+3)+\log (1+3+5)+...+...+\log (1+3+5+7+...+19)]$ - $2[\log 1+\log 2+\log 3+...+\log 7]=p+qa+rb$ 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. Cannot be determined

Answer 1

The correct option is B $26$

$\log 2=a$ and $\log 3=b$

$[\log \left(1\right)+\log (1+3)+\log (1+3+5)+......+\log (1+3+......+19)]$

$=[\log \left(1^2\right)+\log \left(2^2\right)+\log \left(3^2\right)+......+\log \left({10}^2\right)]$

$=[2\log 1+2\log 2+2\log 3+......+2\log 10]$

$=2[\log 1+\log 2+........+\log 10]$

$2[\log 1+\log 2+........+\log 10]-2[\log 1+\log 2+........+\log 7]$

$=2[\log 8+\log 9+\log 10]$

$=2[{\log }_{10}{2}^3+{\log }_{10}{3}^2+{\log }_{10}10]$

$=2[3a+2b+1]=2+6a+4b$

$\therefore p=2;q=6;r=4$

$p+2q+3r=2+12+12=26$