I'm having a hard time with this exponential equation, I'm sure I'm doing some kind of a "minor" mistake again somewhere along the way. Your help is very much appreciated.

$$4\cdot4^{2x}-9\cdot 4^x+2=0$$

*Answers given are: $0,5; -1$.


Let $y=4^x$. The equation $$4\cdot4^{2x}-9\cdot 4^x+2=0$$ $$4\cdot(4^x)^2-9\cdot 4^x+2=0$$ becomes a quadratic $$4y^2-9y+2=0,$$ which we can solve with the quadratic formula. Then, once you've found the two values for $y$, solve for $x$ using $y=4^x$.

  • $\begingroup$ Oh! I forgot about that. Thank you again Zev! $\endgroup$ – BeatShot Aug 27 '11 at 16:13
  • $\begingroup$ No problem, glad to help! $\endgroup$ – Zev Chonoles Aug 27 '11 at 16:13

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