How do you write $\ln 100 = 4.61$in exponential form?

Hint:In order to determine the value of the above question in exponential form ,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 form your answer accordingly.

Complete step by step solution:
To solve the given question, we must know the properties of logarithms and with the help of then we are going to rewrite our question.

Recall that $\ln $is nothing but logarithm having base $e$.We can rewrite our expression as
${\log _e}100 = 4.61$

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 ${\log _e}100 = 4.61$and if compare this with ${\log _b}x = y$we get
$
b = e \\
y = 4.61 \\
x = 100 \\
$
So, ${\log _e}100 = 4.61$is equivalent to ${e^{4.61}} = 100$

Hence the exponential form of $\ln 100 = 4.61$will be equivalent to ${e^{4.61}} = 100$.

Therefore, our required answer is ${e^{4.61}} = 100$.

Note: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.

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 x2 × y3 can't be disentangled on the grounds that the bases (th$n\log m = \log {m^n}$e x and y) are not the equivalent.