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Question: How do you solve \( \ln \left( {y - 1} \right) - \ln 2 = x + \ln \left( x \right) \) ?...

How do you solve ln⁡(y−1)−ln⁡2=x+ln⁡(x)\ln \left( {y - 1} \right) - \ln 2 = x + \ln \left( x \right) ?

Explanation

Solution

Hint : In the given question, we need to solve the equation given to us in the problem ln⁡(y−1)−ln⁡2=x+ln⁡(x)\ln \left( {y - 1} \right) - \ln 2 = x + \ln \left( x \right) . We are only given one equation in two variables. So, we have to express one variable in terms of another using the method of transposition. Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease.

Complete step by step solution:
We would use the method of transposition to find the value of one variable in terms of another ln⁡(y−1)−ln⁡2=x+ln⁡(x)\ln \left( {y - 1} \right) - \ln 2 = x + \ln \left( x \right) using the transposition method. Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together and isolating the variable or the unknown term in order to simplify the equation and finding value of the required parameter.
Now, In order to find the value of x, we need to isolate x from the rest of the parameters.
So, ln⁡(y−1)−ln⁡2=x+ln⁡(x)\ln \left( {y - 1} \right) - \ln 2 = x + \ln \left( x \right)
Now, using the logarithm property log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left( {\dfrac{a}{b}} \right) , we get,
⇒ln⁡(y−12)=x+ln⁡(x)\Rightarrow \ln \left( {\dfrac{{y - 1}}{2}} \right) = x + \ln \left( x \right)
Now, taking anti logarithm on both sides of the equation, we get,
⇒eln⁡(y−12)=ex+ln⁡(x)\Rightarrow {e^{\ln \left( {\dfrac{{y - 1}}{2}} \right)}} = {e^{x + \ln \left( x \right)}}
Now, we know that eln⁡a=a{e^{\ln a}} = a . So, we get,
⇒(y−12)=ex+ln⁡(x)\Rightarrow \left( {\dfrac{{y - 1}}{2}} \right) = {e^{x + \ln \left( x \right)}}
Now, we use the exponential property em+n=em×en{e^{m + n}} = {e^m} \times {e^n} .
⇒(y−12)=ex×eln⁡x\Rightarrow \left( {\dfrac{{y - 1}}{2}} \right) = {e^x} \times {e^{\ln x}}
Again using eln⁡a=a{e^{\ln a}} = a , we get,
⇒(y−12)=ex(x)\Rightarrow \left( {\dfrac{{y - 1}}{2}} \right) = {e^x}\left( x \right)
Simplifying further, we get,
⇒y−12=xex\Rightarrow \dfrac{{y - 1}}{2} = x{e^x}
Now, isolating y in order to find its value in terms of x, we get,
⇒y−1=2xex\Rightarrow y - 1 = 2x{e^x}
Adding 11 to both sides of the equation, we get,
⇒y=2xex+1\Rightarrow y = 2x{e^x} + 1
Hence, the equation ln⁡(y−1)−ln⁡2=x+ln⁡(x)\ln \left( {y - 1} \right) - \ln 2 = x + \ln \left( x \right) can be simplified as y=2xex+1y = 2x{e^x} + 1 using the logarithm properties and simplification.
So, the correct answer is “ y=2xex+1y = 2x{e^x} + 1 ”.

Note : If we add, subtract, multiply or divide by the same number on both sides of a given algebraic equation, then both sides will remain equal. The given problem deals with an algebraic equation. There is no fixed way of solving a given algebraic equation. Algebraic equations can be solved in various ways. Linear equations in one variable can be solved by the transposition method with ease.