1953 AMC 12 Problem 47

Question

If

x

is greater than zero, then the correct relationship is:

\log(1+x)=\dfrac{x}{1+x}

\log(1+x)\lt\dfrac{x}{1+x}

\log(1+x)\gt x

\log(1+x)\lt x

none of these

Compare

1+x

with an exponential function for

x\gt0

The standard inequality

\ln(1+x)\lt x

is even stronger for common logarithms

For

x\gt0,

the standard exponential inequality gives

1+x\lt e^x.

Taking natural logarithms yields

\ln(1+x)\lt x.

If

\log

denotes the common logarithm, then

\begin{aligned} \log(1+x)&=\frac{\ln(1+x)}{\ln10}\\ &\lt\ln(1+x), \end{aligned}

so the same listed inequality holds.

Thus, the correct answer is D.

Problem 47 in other years