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Question: How do you solve \(\ln \left( x \right)+\ln \left( x-1 \right)=1\)...

How do you solve ln(x)+ln(x1)=1\ln \left( x \right)+\ln \left( x-1 \right)=1

Explanation

Solution

To solve the given equation we will first use the property of log log(MN)=logM+logN\log \left( MN \right)=\log M+ \log N . Now if we use the definition of log we will get a quadratic equation in x. We will solve the quadratic equation by using the formula b±b24ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} . Hence we will get the solution of the given equation.

Complete step-by-step solution:
Let us first understand the concept of logarithm.
Logarithm of a number x is the exponent to which base must be raised to produce the given number.
Hence let us say we have 102=100{{10}^{2}}=100
In logarithm form this can be written as log10(100)=2{{\log }_{10}}\left( 100 \right)=2 in which 10 is called the base of logarithm.
Now if the base of logarithm is taken as e then the log is called natural logarithm and is written as ln.
Now let us understand three basic properties of logarithm.
logb(MN)=logb(M)+logb(N){{\log }_{b}}(MN)={{\log }_{b}}\left( M \right)+{{\log }_{b}}\left( N \right)
logb(MN)=logb(M)logb(N){{\log }_{b}}\left( \dfrac{M}{N} \right)={{\log }_{b}}\left( M \right)-{{\log }_{b}}\left( N \right)
And logb(M)p=plogb(M){{\log }_{b}}{{\left( M \right)}^{p}}=p{{\log }_{b}}\left( M \right)
Now consider the given expression ln(x)+ln(x1)=1\ln \left( x \right)+\ln \left( x-1 \right)=1
Now we know that logb(MN)=logb(M)+logb(N){{\log }_{b}}(MN)={{\log }_{b}}\left( M \right)+{{\log }_{b}}\left( N \right) .
Hence using this we get,
ln[(x)(x1)]=1\Rightarrow \ln \left[ \left( x \right)\left( x-1 \right) \right]=1
Now we know ln is nothing but the logarithm of base e. Hence using the definition of logarithm we get,
x(x1)=e1\Rightarrow x\left( x-1 \right)={{e}^{1}}
x2xe=0\Rightarrow {{x}^{2}}-x-e=0
Now this is a quadratic equation of the form ax2+bx+c=0a{{x}^{2}}+bx+c=0 where a = 1, b = - 1 and c = - e.
We know that the roots of the quadratic equation are given by b±b24ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} . Substituting the values of a, b and c in the formula we get,
x=(1)±(1)24(1)(e)2(1) x=1±1+4e2 \begin{aligned} & \Rightarrow x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -e \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{1\pm \sqrt{1+4e}}{2} \\\ \end{aligned}
Hence the solution of the given equation x=1±1+4e2x=\dfrac{1\pm \sqrt{1+4e}}{2} .

Note: Note that logarithm with natural base is nothing but the inverse function of e. Hence eln(x)=x{{e}^{\ln \left( x \right)}}=x . Similarly we can say e is the inverse function of log and hence ln(ex)=x\ln \left( {{e}^{x}} \right)=x. Note that ln(e)=1\ln \left( e \right)=1. Hence while solving logarithmic equations we can also use e function to eliminate log.