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# Exponents

0
146
5

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

Apr 15, 2022

#2
+2540
+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
+124697
+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

Apr 15, 2022
#3
+2540
0

Wait... isn't it $$27^n$$...

BuilderBoi  Apr 15, 2022
#5
+124697
0

You are correct  !!!!

Disregard mine.....!!!!

CPhill  Apr 15, 2022
#2
+2540
+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 $$\color{brown}\boxed{n=-1.2}$$

BuilderBoi Apr 15, 2022
#4
+124697
+1