Question

If ${\log }_{8}a+{\log }_4{b}^2=5$, ${\log }_{8}b+{\log }_4{a}^2=7$, then the value of $(a\cdot b)$ is equal to

• A. $128$
• B. $256$
• C. $512$
• D. $1024$

Answer 1

The correct option is C $512$

${\log }_{8}a+{\log }_4{b}^2=5{\log }_{8}b+{\log }_4{a}^2=7\text{Taking base 2}\frac{{\log }_{2}a}{{\log }_{2}8}+\frac{2{\log }_{2}b}{{\log }_{2}4}=5...\left(i\right)\frac{{\log }_{2}b}{{\log }_{2}8}+\frac{2{\log }_{2}a}{{\log }_{2}4}=7...\left(ii\right)\text{Solving eq.(i)}\frac{{\log }_{2}a}{3}+\frac{2{\log }_{2}b}{2}=5\frac{2{\log }_{2}a}{6}+\frac{6{\log }_{2}b}{6}=52{\log }_{2}a+6{\log }_{2}b=302{\log }_{2}a=30-6{\log }_{2}b...\left(iii\right)\text{Solving eq. (ii) we get}\frac{{\log }_{2}b}{3}+\frac{2{\log }_{2}a}{2}=72{\log }_{2}b+6{\log }_{2}a=42\text{Substituting from eq. (iii), we get}2{\log }_{2}b+3(30-6{\log }_{2}b)=422\log b+90-18\log b=42-16\log b=-48{\log }_{2}b=3b=8\text{Substitute b = 8 in eq. (iii), we get}2{\log }_{2}a=30-6{\log }_{2}b=30-6\times 3=12{\log }_{2}a=6a=2^6=64b=8a.b=64\cdot 8=512$