Question

If x is greater than zero, then which of the following is true
$\begin{align}
  & a)log(1+x)=\dfrac{x}{1+x} \\
 & b)\log (1+x)> x \\
 & c)\log (1+x)< \dfrac{x}{1+x} \\
 & d)None\text{ }of\text{ }these \\
 & e)\text{ }\log (1+x)< x \\
\end{align}$

Answer 1

Hint: Now we have x > 0 and we know by definition of log that $\log a=b$ means $a={{10}^{b}}$. Hence we will make use of this to verify each option and find a value of x to show the options are incorrect.
We know that $\log x > 0$ if x > 1 and $\log x < 0$ is x < 1. We will use this as well as substitute values of x to check if the options are correct or incorrect.

Complete step by step answer:
Now let us check the first option $\log (1+x)=\dfrac{x}{1+x}$
Let us check the equation at x = 1
LHS is $\log (1+1)=\log 2=0.3$
RHS is $\dfrac{1}{1+1}=\dfrac{1}{2}=0.5$
Here we have $\log (1+x)<\dfrac{x}{1+x}$
Hence we can see that for x = 1 the equation $\log (1+x)=\dfrac{x}{1+x}$ fails.
Hence option a is not true.
Now let us check option b $\log (1+x)>x$
Now we know that for any positive number x
$x+1<{{10}^{x}}$
Now let us take log on both side of this equation
$\log (x+1)<\log {{10}^{x}}$
Now we have a property of log which says $\log {{a}^{b}}=b\log a$
Hence we get $\log (x+1) < x\log (10)$
But the value of log 10 is 1. Substituting this we get
$\log (x+1) < x$
Hence option b is true.
Now let us check option c which says $\log (1+x)<\dfrac{x}{1+x}$
Now we know log10 is 1. So let us substitute x = 9 and check if the relation holds
LHS $\log (1+9)=\log 10=1$
RHS = $\dfrac{9}{1+9}=\dfrac{9}{10}=0.9$
Hence the relation is not true for x = 9.
Hence option c is incorrect.
Now option d is none of these
Since we already got option b as correct option d is incorrect
Now option e is $\log (1+x) < x$
Now also since option b is the correct option e can’t hold true.

So, the correct answer is “Option b”.

Note: Now here we solved by checking each option while solving we got $\log (1+x)<\dfrac{x}{1+x}$ at x = 1. Hence we might think that option c is correct while for x = 9 we get $\log (1+x)>\dfrac{x}{1+x}$ . Hence here we have to intuitively select the correct value to disprove. Now to get the intuition we will try to understand the nature of the functions. Now we know that $\dfrac{x}{1+x}$ will always be between 0 to 1 while $\log (1+x)$ can give values beyond 1. Hence we already know that the inequality $\log (1+x)<\dfrac{x}{1+x}$ won’t be true.