Question

How do you solve \[\ln x + \ln (x - 2) = 1\]?

Answer 1

Hint:To solve the expression for logarithmic terms, we should know the logarithmic properties like multiplication division and when we remove the log term then the equation R.H.S will get the term of “e” and the whole expression will goes to the power the this term “e”, value of this function is 2.71828, if needed then we can solve it by using this value.

Complete step by step solution:
The given question needs to solve the expression \[\ln x + \ln (x - 2) = 1\]
Using the property of subtraction in logarithm we can say:
\[ \Rightarrow \ln a + \ln b = \ln a \times b\]
Using this property in our question we can write the given expression as:
\[
\Rightarrow \ln x + \ln (x - 2) = 1 \\
\Rightarrow \ln x \times (x - 2) = 1 \\
\]
Now using the second property which is when you remove the logarithm function then the expression on the second hand of equation will goes to the power of “e”, this property can be written as:
\[
\Rightarrow \ln x = y \\
then \\
\Rightarrow x = {e^y} \\
\]
Using this property in our question we can solve as:
\[
\Rightarrow \ln x \times (x - 2) = 1 \\
\Rightarrow x \times (x - 2) = {e^1} \\
\Rightarrow {x^2} - 2x = {e^1} \\
\Rightarrow {x^2} - 2x - 2.78 = 0 \\

\Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt { - {2^2} - 4 \times 1 \times ( - 2.78)} }}{{2 \times 1}} = \dfrac{{2 \pm \sqrt {4 + 10.42} }}{2} \\
\Rightarrow x = \dfrac{{2 \pm 3.79}}{2} = \dfrac{{5.79}}{2},\dfrac{{ - 1.79}}{2} = 2.895, - 0.895
\\
\]
Hence we obtained the final answer for the given expression.
Formulae Used:
\[ \Rightarrow \ln a - \ln b = \ln \dfrac{a}{b}\]
\[
\Rightarrow \ln x = y \\
then \\
\Rightarrow x = {e^y} \\
\]
Additional Information: The given expression needs to be solved as steps are used above, these are standard steps and need to be used for getting the final answer, the final answer can be cross checked by putting the value of the variable in the equation given in question.

Note: Here the given expression can also be solved by differentiating the equation, on differentiation the log term will be simplified to the normal integer and variable, here using differentiation the final answer would not be changed, you will obtain the same result as we get here.