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Question

Question: How do you solve \[\log \left( x \right) + \log \left( {x + 1} \right) = \log \left( {12} \right)\...

How do you solve log(x)+log(x+1)=log(12)\log \left( x \right) + \log \left( {x + 1} \right) = \log \left( {12} \right)?

Explanation

Solution

In the given question, we need to find the value of x as per the equation log(x)+log(x+1)=log(12)\log \left( x \right) + \log \left( {x + 1} \right) = \log \left( {12} \right). Apply the law of logarithms to solve this equation which isloga+logb=logab\log a + \log b = \log ab,further apply log properties to solve this equation.

Complete step by step answer:
Let us write the given equation
log(x)+log(x+1)=log(12)\log \left( x \right) + \log \left( {x + 1} \right) = \log \left( {12} \right)
Here we can see in the equation that all the terms are with respect to log function
1=log(x)1 = \log \left( x \right)
If the base is 10, then we can write the above term as
101=x{10^1} = x
This implies x=10
Hence,
1=log101 = \log 10
Now let us rewrite the given equation with respect to the terms implied
logx+logx+log10+log12\log x + \log x + \log 10 + \log 12
As the general rule of the logarithm is
logabc=loga+logb+logc\log a \cdot b \cdot c = \log a + \log b + \log c
Hence applying the general rule to the equation, we get
logx+logx+log10=logxx10\log x + \log x + \log 10 = \log x \cdot x \cdot 10
Therefore, after simplifying we get
log10x2=log12\log 10{x^2} = \log 12
This implies that if the logs are equal, then the numbers are equal
Hence, by basic definition of logarithms:
10x2=1210{x^2} = 12
x2=65{x^2} = \dfrac{6}{5}
Therefore, the value of x is
x=65x = \sqrt {\dfrac{6}{5}}

Additional information: Rules of Logarithms
The logarithm of a positive real number can be negative, zero or positive.
Logarithmic values of a given number are different for different bases.
Logarithms to the base a 10 are referred to as common logarithms. When a logarithm is written without a subscript base, we assume the base to be 10.
The logarithmic value of a negative number is imaginary and the logarithm of any positive number to the same base is equal to 1.
a1=alogaa=1{a^1} = a \Rightarrow {\log _a}a = 1
The logarithm of 1 to any finite non-zero base is zero.
a0=1loga1=0{a^0} = 1 \Rightarrow {\log _a}1 = 0
Formula used:
loga+logb=logab\log a + \log b = \log ab

Note: The key point to find the value of x in the given equation is that applying the formula loga+logb=logab\log a + \log b = \log ab, when the equation consists of two variables and hence by applying the logarithmic properties, we can get the value of x