I don't remember what I supposed to do in this situation...I know that it's necessary transform both sides of the equation in the same base. However, what I need to do when i have a 0?
My equation: $e^{2x^3 - 6x^2 + 3} = 0$
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7Is $e^y$ ever $0$? – Daniel Fischer Jun 20 '14 at 18:45
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yes Daniel...i don't know what should I do – EricHideki Jun 20 '14 at 18:49
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it's wrong do this: e^(2x^3-6x^2+3) = e^0 isn't it? – EricHideki Jun 20 '14 at 18:50
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1Yes, $e^0 = 1$. Since $e^y \neq 0$ for all $y$, the equation has no solution. – Daniel Fischer Jun 20 '14 at 18:52
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You can consider the point $x$ at $-\infty$. – Mhenni Benghorbal Jun 20 '14 at 19:06
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We have to know that $e^{\mathrm{something}}$ is never zero, so that doesn't have any solution. It is never negative, also. But, if you had $$e^{\mathrm{something}} = c > 0$$ we can take $\ln$ on both sides and get: $$\begin{align}\ln e^{\mathrm{something}} &= \ln c \\ \mathrm{something} &= \ln c\end{align}$$ That last equation you should be able to solve, normally. Don't be afraid of $\ln c$, it is just a number. Ok?
Ivo Terek
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Thanks Ivo..I've been concluding this also...so...just to confirm: e^something = not exist (right?) – EricHideki Jun 20 '14 at 18:55
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No. Actually, I meant that $e^{\mathrm{something}} \neq 0$, whatever something may be. In your exercise, you have an exponential equal to zero, but that is impossible, hence, no solution. The rest of my answer is how to go at the problem, if the exponential is equal to something positive, meaning that a solution does exist. – Ivo Terek Jun 20 '14 at 18:57