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Find the value of y such that 9^y = 3^5/27^3.

 Oct 29, 2021
 #1
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+1

3^5/27^3

3^5  / ( 3^3)^3

3^5 / 3^9

3^-4   for the right side

 

for the left side

(3^2)^y 

 

 

so    3^(2y) = 3^(-4)      can you finish from here (Hint: equate the exponents)

 Oct 29, 2021
 #2
avatar+390 
+2

Hi Guest,

 

\(9^y=\frac{3^5}{27^3}\)

\(\mbox {Calculate}\)

\(9^y=\frac{243}{19683}\)

\(\mbox {Now divide with 243}\)

\(9^y=\frac{1}{81}\)

\(\mbox {Write the expression in exponential form with the base of 3}\)

\({3}^{2y}=\frac{1}{81}\)

\(\mbox {Write the expression in exponential form with the base of 3}\)

\({3}^{2y}={3}^{-4}\)

\(\mbox {Since the bases are the same, set the exponents equal}\)

\(2y=-4\)

\(\mbox {Divide both sides of the equation by 2}\)

\(y=-2\)

 

Here's your answer.

 

smiley !

 Oct 31, 2021
 #3
avatar+115335 
0

Both of you have answered well but guest's answer is the better one.

 Oct 31, 2021

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