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How do you solve $\ln x-\ln 3=4$?

Answer
VerifiedVerified
541.5k+ views
Hint: In this question, we are given an equation in terms of logarithmic function and we need to solve it to find the value of x. For this we will first use the property of logarithm according to which ${{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right)$ to simplify the left side. Then we will use the property that ${{\log }_{b}}x=a\Rightarrow x={{b}^{a}}$ to form an equation in terms of x which we will solve to get the value of x. We will use the value of e is 2.718

Complete step-by-step answer:
Here we are given the equation as $\ln x-\ln 3=4$.
We need to solve it to find the value of x ${{\log }_{e}}x-{{\log }_{e}}3=4$.
As we know that, ln represent the natural log that is the log with base e. So we can write this equation as.
As we know from the properties of logarithm that ${{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right)$. So let us use it on the left side of the equation we get ${{\log }_{e}}\left( \dfrac{x}{3} \right)=4$.
Now let us further simplify the equation. We know from the property of logarithm and exponents that if ${{\log }_{a}}x=b$ then this means $x={{a}^{b}}$. So let us use this here.
We have ${{\log }_{e}}\left( \dfrac{x}{3} \right)=4$ so we get $\dfrac{x}{3}={{e}^{4}}$.
Now let us solve this to get the value of x.
For removing 3 in the denominator let us multiply both sides of the equation by 3 we get $\dfrac{x}{3}\times 3=3\times {{e}^{4}}$.
Simplifying we get $x=3{{e}^{4}}$.
This is our required value of x where e = 2.718.
Let us calculate the actual value as well. Taking e = 2.718 we get,
$x=3{{\left( 2.718 \right)}^{4}}\Rightarrow 3\times 54.57551\Rightarrow 163.8$.
Hence the required value of x is 1638 which will satisfy the given equation.

Note: Students should take care of the signs while using logarithm property. Note that if we are adding two log terms, angles are multiplied whereas if terms are subtracted then angles are divided. Students can leave their answers in the form of e also.