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I have the following equation:

$$y=\exp(C_1-C_2\ln x)$$

where $C_1$ and $C_2$ are constants.

I believe that the following is correct:

$$y=\exp C_1\exp (-C_2\ln x)$$

$\exp C_1$ is just a constant. I know that the exponent of the natural logarithm of $x$ is just $x$. But I'm not sure what to do with $C_2$.

Robert Shore
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rdemyan
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    This is

    $$y=\frac{e^{C_1}}{x^{C_2}}.$$

     – Robert Shore Apr 08 '23 at 02:48
  • Notice that (1) $\exp C_1$ is itself a positive constant (an arbitrary one if $C_1$ can take on any real value), and (2) $\exp(-C_2 \ln x) = (e^{\ln x})^{-C_2}=x^{-C_2}$, where the expression on the right-hand side is restricted to $\operatorname{dom} \ln = (0, \infty)$. – Travis Willse Apr 08 '23 at 02:57
  • Note that $\exp(\ln x) = x$, $\exp(-x) = 1/\exp(x)$ and $\exp(ab) = \left(\exp(a)\right)^b$ – D S Apr 08 '23 at 03:50

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