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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.

Guest Jul 25, 2018
edited by Guest  Jul 25, 2018
 #1
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 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

Guest Jul 25, 2018
 #2
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Find the value of $y$ such that $9^y = \dfrac{3^5\cdot 9^6}{27^3}$.

\(\begin{array}{|rcll|} \hline 9^y &=& \dfrac{3^5\cdot 9^6}{27^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{(3\cdot 9)^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{3^39^3} \\\\ 9^y &=& 3^{5-3}\cdot 9^{6-3} \\ 9^y &=& 3^{2}\cdot 9^{3} \\ 9^y &=& 9\cdot 9^{3} \\ 9^y &=& 9^1\cdot 9^{3} \\ 9^y &=& 9^{3+1} \\ 9^{{\color{red}y}} &=& 9^{{\color{red}4}} \\\\ \mathbf{y} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

 

laugh

heureka  Jul 25, 2018

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