Question

How do you solve $2\ln x=1$ ?

Answer 1

Hint: To solve this question first divide the entire equation on both sides with 2 and now that ln x is isolated, apply $e$ on both sides, the RHS and the LHS of the expression. Since we know that ${{e}^{\ln x}}=x$ , simplify the terms and then find the value of x.

Complete step by step solution:
The given logarithmic expression is $2\ln x=1$
Now let us start solving it by dividing the entire expression with 2.
$\Rightarrow \ln x=\dfrac{1}{2}$
Now we have isolated the lnx. The next step is to isolate the x, we need to apply $e$ on both sides of the expression.
The number, e is a mathematical constant and irrational number which is numerically equal to 2.71828.
The expansion of the function is,
$e$ is a positive number so when raised to any power yields a positive value.
Thus, the domain of ${{e}^{x}}$ is a set of real numbers and the exponential function ranges from zero to infinity.
$\Rightarrow {{e}^{\ln x}}={{e}^{\dfrac{1}{2}}}$
We know that ${{e}^{\ln x}}=x$
Hence the expression will be simplified as,
$\Rightarrow x={{e}^{\dfrac{1}{2}}}$
This value is approximately equal to 1.648

Hence, the solution for $2\ln x=1$ is $x={{e}^{\dfrac{1}{2}}}$ which is approximately equal to the value 1.648

Note: lnx is nothing but ${{\log }_{e}}x$ .It is a logarithm with the base $e$ . Now, the domain of ${{e}^{x}}$ is a set of real numbers and the exponential function ranges from zero to infinity. The derivative of the exponential function is always greater than zero. Hence the functions do not have any critical points which make it continuous throughout its domain. The function ${{e}^{x}}$ can be expressed in series form as:
$\Rightarrow {{e}^{x}}=\displaystyle \lim_{n \to \infty }{{\left( \dfrac{1+x}{n} \right)}^{n}}$
The value of limit exists for every number that belongs to the set of real numbers or the domain of ${{e}^{x}}$