Need help with 2/3*n=1. Find n.

Is that \(\displaystyle \L\;\frac{2}{3^{n}}\) or \(\displaystyle \L\;[\frac{2}{3}]^{n}\)?

The one on the right is very easy, since \(\displaystyle \L\;a^{x} = 1\), for a > 0 requires x = 0. The one on the left is a little trickier. Are you ready for logarithms?
 
Since you posted this problem in the Pre-Algebra forum, I wonder if your problem is

(2/3)*n = 1

If this is what you meant, then we want to get 1n, or just n, on the left side. Since n has been multiplied by 2/3, we can multiply both sides by 3/2:

(3/2)*(2/3)*n = (3/2)*1

(3/2) and (2/3) are reciprocals; when you multiply a number by its reciprocal, the product is 1:

1*n = (3/2)*1
n = 3/2

Check. Substitute 3/2 for "n" in the original equation:

(2/3)*n = 1
(2/3)*(3/2) = 1
1 = 1
It checks.

If this is not what you meant in your problem, please respost clarifying your notation.
 
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