+118723 Let me think about this.
Firstly is this what you mean, or is the -1 tacked onto the end?
$$\left(\frac{1}{3}\right)^8=\left(\frac{-1}{3}\right)^{n-1}$$
if (n-1) is even then
$$\left(\frac{1}{3}\right)^8=\left(\frac{1}{3}\right)^{n-1}$$
n-1=8
n=9
If n-1 is odd it can't work.
okay so n=9 is the only answer in the Real number system.
+118723 Let me think about this.
Firstly is this what you mean, or is the -1 tacked onto the end?
$$\left(\frac{1}{3}\right)^8=\left(\frac{-1}{3}\right)^{n-1}$$
if (n-1) is even then
$$\left(\frac{1}{3}\right)^8=\left(\frac{1}{3}\right)^{n-1}$$
n-1=8
n=9
If n-1 is odd it can't work.
okay so n=9 is the only answer in the Real number system.
