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.

Guest Jun 24, 2019

#1**+3 **

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

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 9^{5} as 3^{10}

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

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

\(x\ =\ \pm243\)

.hectictar 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) = ± [3]^5 = ± 243

CPhill 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?**

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

heureka Jun 24, 2019