Find the value of $y$ such that $9^y = \dfrac{3^5\cdot 9^6}{27^3}$.
Got 7 the first time and 1 the sencond time... Don't know what's wrong...
Thanks in advance.
9^y = (3^5×9^6)/27^3 solve for y
9^y = 6,561 factor both sides
3^2y = 3^8 equate the exponents
2y = 8 divide both sides by 2
y = 8/2
y = 4
+26367 Find the value of $y$ such that $9^y = \dfrac{3^5\cdot 9^6}{27^3}$.
\(\begin{array}{|rcll|} \hline 9^y &=& \dfrac{3^5\cdot 9^6}{27^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{(3\cdot 9)^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{3^39^3} \\\\ 9^y &=& 3^{5-3}\cdot 9^{6-3} \\ 9^y &=& 3^{2}\cdot 9^{3} \\ 9^y &=& 9\cdot 9^{3} \\ 9^y &=& 9^1\cdot 9^{3} \\ 9^y &=& 9^{3+1} \\ 9^{{\color{red}y}} &=& 9^{{\color{red}4}} \\\\ \mathbf{y} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)
