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Using the graph with which the natural logarithm ln is geometrically defined, show that $\ln (1 + x) <x$.

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2 Answers

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First, it can be helpful to have a picture:

In this picture, the purple/blue area is the area below the curve $y=\frac{1}{x}$ from $1$ to about 6.25, i.e. $\ln(1+5.25)$, while the red area is the area under curve $y=1$ over the same range, i.e. which is about $5.25$. The Desmos demos is linked here if you would like to play with this a bit more.

Generalizing this static picture, the red area represents $x$, while the blue area is $\ln(1+x)$. From the picture, it seems clear which is larger.

Being a bit more careful, we can work symbolically, using the definitions directly: for any $x > 0$, we have $$ \ln(x) := \int_{1}^{x} \frac{1}{t} \,\mathrm{d}t. $$ This implies that $$ \ln(1+x) = \int_{1}^{1+x} \frac{1}{t}\, \mathrm{d}t. $$ On the other hand, observe that $$ x = \int_{1}^{1+x} 1 \,\mathrm{d}t; $$ that is, thinking of the integral as an area, we can regard $x$ as the area of a rectangle of height 1 and width $x$, with the lower-left corner at the point $(1,0)$. But if $x > 1$, then we have $\frac{1}{x} < 1$, and so by the monotonicity of the integral, we have $$ \log(1+x) = \int_{1}^{1+x} \frac{1}{t}\, \mathrm{d}t \le \int_{1}^{1+x} 1 \,\mathrm{d}t = x, $$ which is the desired result.

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$f(x)=\ln(1+x)$ is a concave function and $y=x$ is a tangent line to $f$ in $(0,0)$.

Thus, $f(x)\leq x$ for all $x>-1$. Draw it!

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