Question

The ratio when simplified $\frac{2^{{\log }_{2^{1/4}}a}-3^{{\log }_{27}(a^2+1{)}^3}-2a}{7^{4{\log }_{49}a}-a-1}$ gives

• A. $a^2-a-1$
• B. $a^2+a-1$
• C. $a^2-a+1$
• D. $a^2+a+1$

Answer 1

The correct option is D $a^2+a+1$

$\frac{2^{{\log }_{2^{1/4}}a}-3^{{\log }_{27}(a^2+1{)}^3}-2a}{7^{4{\log }_{49}a}-a-1}$

$=\frac{2^{{\log }_2{a}^4}-3^{{\log }_3(a^2+1)}-2a}{7^{{\log }_7{a}^2}-a-1}$

$=\frac{a^4-(a^2+1)-2a}{a^2-a-1}[\because a^{{\log }_{a}b}=b]$

$=\frac{a^4-(a+1{)}^2}{a^2-a-1}$

$=\frac{(a^2-a-1)(a^2+a+1)}{a^2-a-1}=a^2+a+1$

Hence option 'D' is correct choice.