Question
Question: How do you solve \(\ln \left( x \right)+\ln \left( x-1 \right)=1\)...
How do you solve ln(x)+ln(x−1)=1
Solution
To solve the given equation we will first use the property of log log(MN)=logM+logN . 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 2a−b±b2−4ac . 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
In logarithm form this can be written as log10(100)=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)
logb(NM)=logb(M)−logb(N)
And logb(M)p=plogb(M)
Now consider the given expression ln(x)+ln(x−1)=1
Now we know that logb(MN)=logb(M)+logb(N) .
Hence using this we get,
⇒ln[(x)(x−1)]=1
Now we know ln is nothing but the logarithm of base e. Hence using the definition of logarithm we get,
⇒x(x−1)=e1
⇒x2−x−e=0
Now this is a quadratic equation of the form ax2+bx+c=0 where a = 1, b = - 1 and c = - e.
We know that the roots of the quadratic equation are given by 2a−b±b2−4ac . Substituting the values of a, b and c in the formula we get,
⇒x=2(1)−(−1)±(−1)2−4(1)(−e)⇒x=21±1+4e
Hence the solution of the given equation x=21±1+4e .
Note: Note that logarithm with natural base is nothing but the inverse function of e. Hence eln(x)=x . Similarly we can say e is the inverse function of log and hence ln(ex)=x. Note that ln(e)=1. Hence while solving logarithmic equations we can also use e function to eliminate log.