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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
+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

$$x\ =\ \pm\sqrt{3^{10}}$$

$$x\ =\ \pm3^5$$

$$x\ =\ \pm243$$

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Jun 24, 2019
#2
+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)  =  ± ^5  =  ± 243   Jun 24, 2019
#3
+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?

$$\begin{array}{|rcll|} \hline \mathbf{x^{400}} &=& \mathbf{9^{1000}} \quad | \quad \text{Take the 400th root of both sides } \\ x^{\frac{400}{400}} &=& 9^{\frac{1000}{400}} \\ x &=& 9^{\frac{5}{2}} \\ x &=& \sqrt{9^5} \\ \mathbf{ x} &=& \mathbf{\pm 243} \\ \hline \end{array}$$ Jun 24, 2019