Question

The value of $\frac{2^{{\log }_{2}4}-3^{{\log }_{27}{({2a}^{}+1)}^3}+2a}{7^{2{\log }_{49}a}-a+1}$ is equal to

Answer 1

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

$\Rightarrow \frac{2^{2lo{g}_2^2}-3^{\frac{1}{3}lo{g}^3(2a+1)}+2a}{7^{2\times \frac{1}{2}}lo{g}_7^a-a+1}$

$\Rightarrow \frac{4-2a-1+2a}{a-a+1}=\frac{3}{1}=3$