Question

How do you solve \[\ln (2x - 5) - \ln (4x) = 2\]?

Answer 1

Hint: We will first use the fact that \[\ln a - \ln b = \ln \dfrac{b}{a}\] and then, we will make use of the fact that: If \[\ln a = b\], then \[a = {e^b}\]. Thus, we will have an equation in x which can be solved for x.

Complete step by step solution:
We are given that we are required to solve \[\ln (2x - 5) - \ln (4x) = 2\].
Now, since we know that we have a formula with us regarding logarithmic function which states the following:-
\[ \Rightarrow \ln a - \ln b = \ln \dfrac{b}{a}\]
Replacing a by (2x – 5) and b by 4x, we will then obtain the following expression with us:-
\[ \Rightarrow \ln \left( {2x - 5} \right) - \ln (4x) = \ln \dfrac{{2x - 5}}{{4x}}\]
Putting this in the given expression, we will then obtain the following expression with us:-
\[ \Rightarrow \ln \dfrac{{2x - 5}}{{4x}} = 2\]
Now, since we know that we have a formula with us regarding logarithmic function which states the following:-
If \[\ln a = b\], then \[a = {e^b}\].
Using the above mentioned expression, we will then obtain the following expression with us:-
\[ \Rightarrow \dfrac{{2x - 5}}{{4x}} = {e^2}\]
Cross – multiplying the above expression, we will then obtain the following expression with us:-
\[ \Rightarrow 2x - 5 = 4x{e^2}\]
Taking \[4x{e^2}\] from addition in the right hand side to subtraction in the left hand side, we will then obtain the following equation with us:-
\[ \Rightarrow 2x - 5 - 4x{e^2} = 0\]
Taking 5 from subtraction in the left hand side to addition in the right hand side, we will then obtain the following equation with us:-
\[ \Rightarrow 2x - 4x{e^2} = 5\]
Taking x common from the left hand side, we will then obtain the following equation with us:-
\[ \Rightarrow x\left( {2 - 4{e^2}} \right) = 5\]
Taking \[\left( {2 - 4{e^2}} \right)\] from multiplication in the left hand side to division in the right hand side, we will then obtain the following equation with us:-
\[ \Rightarrow x = \dfrac{5}{{2 - 4{e^2}}}\]
Thus, we have the required answer.

Note: The students must note that there are some rules and regulations before using the formulas mentioned in the hint and used in the solution given as follows:
\[\ln a - \ln b = \ln \dfrac{b}{a}\] is true for all positive real numbers a and b because logarithmic of negative or zero is not defined.
If \[\ln a = b\], then \[a = {e^b}\] is true for all positive real numbers because logarithmic of negative or zero is not defined but b can take any value from real numbers.