Question

Given $lo{g}_{2}a=p,lo{g}_{4}b=p^2$ and $lo{g}_{c^2}\left(8\right)=\frac{2}{p^3+1}$. If $lo{g}_2\left(\frac{c^8}{a{b}^2}\right)=(\alpha p^3-\beta p^2-\gamma p+\delta )$ where $\alpha ,beta,\gamma ,\delta ,\in N$, then find the value of $(\alpha +\beta +\gamma +\delta )$.

Answer 1

${\log }_{2}a=p$ ......... $\left(i\right)$
${\log }_{4}b=p^2$
$\frac{{\log }_{2}b}{{\log }_{2}4}=p^2\Rightarrow \frac{{\log }_{2}b}{2{\log }_{2}2}=p^2$
${\log }_{2}b=2p^2$ ......... $\left(ii\right)$
${\log }_{c^2}8=\frac{2}{p^3+1}$
$\frac{{\log }_{2}8}{{\log }_2{c}^2}=\frac{2}{p^3+1}$
$\frac{{\log }_{2}8}{2{\log }_{2}c}=\frac{2}{p^3+1}$
$\frac{{\log }_{2}8}{{\log }_{2}c}=\frac{4}{p^3+1}$
$\frac{{\log }_{2}c}{{\log }_{2}8}=\frac{p^3+1}{4}$
${\log }_{8}c=\frac{p^3+1}{4}$
$\frac{{\log }_{2}c}{{\log }_{2}8}=\frac{p^3+1}{4}\Rightarrow \frac{{\log }_{2}c}{3{\log }_{2}2}=\frac{p^3+1}{4}$
${\log }_{2}c=\frac{3p^3+3}{4}$ ......... $\left(iii\right)$
${\log }_2\left(\frac{c^8}{{ab}^2}\right)=\alpha p^3-\beta p^2-\gamma p+\delta$
${\log }_2{c}^8-{\log }_2{ab}^2=\alpha p^3-\beta p^2-\gamma p+\delta$
$8{\log }_{2}c-{\log }_{2}a-2{\log }_{2}b=\alpha p^3-\beta p^2-\gamma p+\delta$
$6p^3+6-p-4p^2=\alpha p^3-\beta p^2-\gamma p+\delta$

By comparing the coefficients, we get

$\alpha =6,\beta =4,\gamma =1,\delta =6$

$\therefore \alpha +\beta +\gamma +\delta =17$