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There are two numbers whose 400th powers are equal to 9^1000. In other words, there are two numbers that can replace x in the equation x^400 = 9^1000 making the equation true. What are those numbers? Explain the process by which you got your answer.

 Jun 24, 2019
 #1
avatar+8852 
+4

\(x^{400}\ =\ 9^{1000}\)

                              Take the  200th root of both sides of the equation.

\(x^{\frac{400}{200}}\ =\ 9^{\frac{1000}{200}}\)

 

\(x^{2}\ =\ 9^{5}\)

                              Take the ± sqrt of both sides. (We could have taken the 400th root of both sides to start with.)

\(x\ =\ \pm\sqrt{9^5}\)

                              Rewrite  95  as  310

\(x\ =\ \pm\sqrt{3^{10}}\)

 

\(x\ =\ \pm3^5\)

 

\(x\ =\ \pm243\)

.
 Jun 24, 2019
 #2
avatar+106539 
+2

Well....since it looks as though we're all at the same picnic.....here's my attempt at a weak answer

 

x^400 = 9^1000

 

Take the GCF of 400, 1000  = 200  ....so we can write

 

(x^2)^200 = (9^5)^200        take the 200th root of both sides

 

(x^2) = 9^5      and we can write

 

(x^2) = (3^2)^5

 

x^2  = 3^10      take both roots

 

x = ± √[3^10]  =   ± [ 3] ^(10/2)  =  ± [3]^5  =  ± 243

 

 

cool cool cool

 Jun 24, 2019
 #3
avatar+23897 
+2

There are two numbers whose 400th powers are equal to 9^1000.

In other words, there are two numbers that can replace x in the equation \(x^{400} = 9^{1000}\) making the equation true.
What are those numbers?

 

\(\begin{array}{|rcll|} \hline \mathbf{x^{400}} &=& \mathbf{9^{1000}} \quad | \quad \text{Take the 400th root of both sides } \\ x^{\frac{400}{400}} &=& 9^{\frac{1000}{400}} \\ x &=& 9^{\frac{5}{2}} \\ x &=& \sqrt{9^5} \\ \mathbf{ x} &=& \mathbf{\pm 243} \\ \hline \end{array}\)

 

laugh

 Jun 24, 2019

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