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For what value of n are (1/9)^3 and 9^n*27^n equal?

 Apr 15, 2022

Best Answer 

 #2
avatar+1384 
+1

We can write this as an equation: \({1 \over 9}^3 = 9^n \times 27^n\)

 

We can rewrite these in powers of 3 as: \(3^{-6} =3^{2n} \times 3^{3n}\)

 

Because the bases are the same, we can rewrite this as an equation with n: \(-6 = 5n\)

 

Simplifying, we find \(\color{brown}\boxed{n=-1.2}\)

 Apr 15, 2022
 #1
avatar+122390 
+1

(1/9)^3  =  27 * 9^n

 

9^(-3)  = 3^3 * 9^n

 

(3^2)^(-3)  = 3^3 * ( 3^2)^n

 

3^ (-6) = 3^3 * 3^(2n)

 

3 ^ (-6)  =  3^ ( 3 + 2n)

 

Solve for the exponents

 

-6  = 3 + 2n

 

-6 - 3  = 2n

 

-9 = 2n

 

n =  -9/2

 

 

cool cool cool

 Apr 15, 2022
 #3
avatar+1384 
0

Wait... isn't it \(27^n\)...

BuilderBoi  Apr 15, 2022
 #5
avatar+122390 
0

You are correct  !!!!

 

Disregard mine.....!!!!

 

cool cool cool

CPhill  Apr 15, 2022
 #2
avatar+1384 
+1
Best Answer

We can write this as an equation: \({1 \over 9}^3 = 9^n \times 27^n\)

 

We can rewrite these in powers of 3 as: \(3^{-6} =3^{2n} \times 3^{3n}\)

 

Because the bases are the same, we can rewrite this as an equation with n: \(-6 = 5n\)

 

Simplifying, we find \(\color{brown}\boxed{n=-1.2}\)

BuilderBoi Apr 15, 2022
 #4
avatar+122390 
+1

NVM.....I mis-read the question...

 

BuilderBoi's answer is  correct  !!!!

 

cool cool cool

 Apr 15, 2022

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