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m) \( \frac{3^{2 x}-3^{x+1}+2}{3^{2 x}+3^{x}-6} \quad 3^{2 x} \)

Ask by Harris Lewis. in South Africa
Mar 18,2025

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Answer

The simplified form of the expression is \( \frac{3^x - 1}{3^x + 3} \), provided that \( 3^x \neq 2 \).

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Bonus Knowledge

To understand this expression better, let’s substitute \( y = 3^x \). The expression then transforms into: \[ \frac{y^2 - 3y + 2}{y^2 + y - 6} \] This change of variables makes the algebra more familiar and manageable. Now, let's simplify! The numerator factors to \( (y-1)(y-2) \), and the denominator factors to \( (y-2)(y+3) \), so we can rewrite it as: \[ \frac{(y-1)(y-2)}{(y-2)(y+3)} \] Here, as long as \( y \neq 2 \) (or \( 3^x \neq 2 \)), we can simplify to: \[ \frac{y-1}{y+3} \] Finally, switching back gives us: \[ \frac{3^x - 1}{3^x + 3} \] This showcases how switching your variable can often lead you to simplifications you might not have seen before!

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