y=2e^-(ln x) + 1 Hey, can someone help me simplify this equation.
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You know that exp(ln(a)) = a? And r.ln(x) = ln(x^r), so -ln(x) = ln(1/x).
Originally Posted by TD! You know that exp(ln(a)) = a? And r.ln(x) = ln(x^r), so -ln(x) = ln(1/x). so... y=2(1/x)+1 y=2/x +1 ?
Originally Posted by Takkun y=2e^-(ln x) + 1 Hey, can someone help me simplify this equation. Or you could look at it this way: $\displaystyle y = 2e^{-ln(x)} + 1 = \frac{2}{e^{ln(x)}} + 1$ and then use TD's trick: $\displaystyle e^{ln(a)} = a$. (It isn't much different, but I prefer to have my exponents positive when possible.) -Dan
Originally Posted by Takkun so... y=2(1/x)+1 y=2/x +1 ? Correct!
Originally Posted by Takkun so... y=2(1/x)+1 y=2/x +1 ? Yes, that's it. -Dan
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