Question

$$lnx^{2}1\div2^{3}+2y; (nx+y)^{2}dx$$

Answer

$$ln(x^(1/4))+2*y;(n*x+y)^2*d*x$$

Solution


Simplify  \({2}^{3}\)  to  \(8\).
\[\begin{aligned}&\ln{({x}^{2})}\times \frac{1}{8}+2y\\&{(nx+y)}^{2}dx\end{aligned}\]
Use Power Rule: \(\log_{b}{{x}^{c}}=c\log_{b}{x}\).
\[\begin{aligned}&\ln{(\sqrt[8]{{x}^{2}})}+2y\\&{(nx+y)}^{2}dx\end{aligned}\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\begin{aligned}&\ln{({x}^{\frac{2}{8}})}+2y\\&{(nx+y)}^{2}dx\end{aligned}\]
Simplify  \(\frac{2}{8}\)  to  \(\frac{1}{4}\).
\[\begin{aligned}&\ln{(\sqrt[4]{x})}+2y\\&{(nx+y)}^{2}dx\end{aligned}\]