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  "body": "\n## Proof\n\nWe show that $\\exp$ is a continuous strictly increasing bijection from $\\R$ onto $(0, \\infty)$, and that $\\ln$ is its inverse.\n\n1. Write $e^y = \\exp(y)$ for the [[exponential function|0RMVS7RXZX]]. It is differentiable by [[its derivative rule|0EX9GFJ0B3]], so it is [[continuous|07ZMX4ANJS]].\n2. It is [[positive|0P9BNBW2W3]] and [[strictly increasing|0KN2V7840E]]. So it maps $\\R$ into $(0, \\infty)$, and it is injective.\n3. By the [[exponential of a sum|0E8HNVBCSP]], $e^y e^{-y} = e^0 = 1$, so $e^{-y} = 1/e^y$.\n4. Let $x \u003e 0$. We find $y$ with $e^y = x$. By the inequality [[$1 + t \\le e^t$|012C7GFF20]], $e^x \\ge 1 + x \u003e x$.\n5. By steps 3 and 4, $e^{-1/x} = \\frac{1}{e^{1/x}} \\le \\frac{1}{1 + 1/x} = \\frac{x}{x + 1} \u003c x$.\n6. By steps 4 and 5, $e^{-1/x} \u003c x \u003c e^x$. Also $-1/x \u003c 0 \u003c x$.\n7. By the [[intermediate value theorem|00KAFBR32E]] on $[-1/x, x]$, there is $y$ with $e^y = x$. By step 2, this $y$ is unique.\n8. So $\\exp$ is a bijection from $\\R$ onto $(0, \\infty)$. For $x \u003e 0$, the [[natural logarithm|0044CPHJBV]] $\\ln x$ is the unique $y$ of step 7. So $\\ln$ is the inverse function of $\\exp$.\n9. For $x \u003e 0$, $e^{\\ln x} = x$ by the definition of $\\ln x$.\n10. For real $y$, the number $e^y$ is positive, and $y$ is a solution of $e^t = e^y$. By uniqueness in step 7, $\\ln(e^y) = y$.\n11. By the theorem on the [[continuity of the inverse|07BY47HG5S]] with $I = \\R$ and $f = \\exp$, the inverse $\\ln$ is continuous and strictly increasing on $\\exp(\\R) = (0, \\infty)$.\n12. By step 10, each real $y$ equals $\\ln(e^y)$, so the range of $\\ln$ is $\\R$.\n\n## Idea\n\nThe exponential function is continuous and strictly increasing. It takes values larger than any given positive $x$ and smaller than $x$, so it hits $x$ exactly once. Its inverse is $\\ln$, and the inverse of a continuous strictly increasing function on an interval is again continuous and strictly increasing.\n",
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