Question

How do you solve $\ln (\ln x) = 1$?

Answer 1

Hint: In order to determine the value of the above question, we will convert the expression into exponential form twice, first to remove the outer $\log $ and then the second $\log $, and to do so use the definition of logarithm that the logarithm of the form ${\log _b}x = y$ is when converted into exponential form is equivalent to ${b^y} = x$, so compare with the given logarithm value with this form and repeat the process one more time, you will get your required answer .

Complete step by step answer:
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
Recall that $\ln $ is nothing but logarithm having base $e$. We can rewrite our expression as
${\log _e}(\ln x) = 1$. Lets convert this into its exponential form to remove ${\log _e}$.
Any logarithmic form ${\log _b}X = y$ when converted into equivalent exponential form results in ${b^y} = X$
So in Our question we are given ${\log _e}(\ln x) = 1$ and if compare this with ${\log _b}x = y$ we get
$
  b = e \\
  y = 1 \\
  X = \ln x \\
 $
\[
   \Rightarrow {\log _e}(\ln x) = 1 \\
   \Rightarrow \ln x = {e^1} \\
   \Rightarrow \ln x = e \\
 \]
Rewriting the above,
\[ \Rightarrow {\log _e}x = e\]
Again, Repeating the same steps to again convert to convert \[{\log _e}x = e\]into exponential form, consider
$
  b = e \\
  y = e \\
  X = x \\
 $
\[
   \Rightarrow {\log _e}x = e \\
   \Rightarrow x = {e^e} \\
 \]
Therefore, the solution to equation $\ln (\ln x) = 1$ is equal to \[x = {e^e}\].

Additional Information:
1. Value of the constant “e” is equal to 2.71828.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number, we actually undo an exponentiation.
3. Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$
6. The above guidelines work just if the bases are the equivalent. For example, the expression ${\log _d}(m) + {\log _b}(n)$ can't be improved, on the grounds that the bases (the "d" and the "b") are not the equivalent, similarly as $x^2$ × $y^3$ can't be disentangled on the grounds that the bases (the x and y) are not the equivalent.

Note: 1. Don’t forgot to cross check your result
2. $\ln $ is known as natural logarithm
3. Logarithm of constant 1 is equal to zero.