Question

How do you find the exact value of $2ln{e^6} - ln{e^5}$?

Answer 1

Hint: In this question, we have to solve the logarithmic expression. For that, the power rule of logarithms is used. Then multiply the coefficients of the logarithm. Now, apply the formula $\ln e = 1$. After that, simplify the expression. The formula of the power rule is as stated below.
$\ln {\left( x \right)^y} = y\ln \left( x \right)$

Complete step by step solution:
In, this question, we want to solve the natural logarithmic expression,
$ \Rightarrow 2ln{e^6} - ln{e^5}$
The natural logarithm is the logarithm to the base e of a number.
First, we will apply the formula of the power rule. That is $\ln {\left( x \right)^y} = y\ln \left( x \right)$.
Let us apply this formula in the above equation.
Therefore,
$ \Rightarrow 2\left( {6\ln e} \right) - 5\ln e$
Let us simplify the expression. Apply the multiplication of the coefficients. The multiplication of 2 and 6 is 12.
Therefore,
$ \Rightarrow 12\ln e - 5\ln e$
Now, we know that $\ln e = 1$
Let us substitute this formula in the above expression.
$ \Rightarrow 12\left( 1 \right) - 5\left( 1 \right)$
Apply the multiplication. The multiplication of any number with 1 gives the number itself. That is the multiplication of 12 and 1 is 12 and also the multiplication of 5 and 1 is 5.
Therefore,
$ \Rightarrow 12 - 5$
Let us apply the subtraction to the above expression. The subtraction of 12 and 5 is 7.
$ \Rightarrow 7$

Hence, the solution of the given natural logarithmic expression is 7.

Note:
Here are some natural logarithm rules and properties.
Product rule: The logarithm of the multiplication of x and y is the sum of the logarithm of x and logarithm of y.
$\ln \left( {x \times y} \right) = \ln \left( x \right) + \ln \left( y \right)$
Quotient rule: The logarithm of the division of x and y is the difference of the logarithm of x and logarithm of y.
$\ln \left( {\dfrac{x}{y}} \right) = \ln \left( x \right) - \ln \left( y \right)$
Power rule: The logarithm of x raised to the power of y is y times the logarithm of x.
$\ln {\left( x \right)^y} = y\ln \left( x \right)$