How do you solve \[\log (x + 2) + \log (x - 1) = 1\]?
Answer
624.6k+ views
Hint: Here in this question, we have to solve the given question. Here we see the word log then the function is logarithmic function so we solve this question we use the logarithmic functions properties. The logarithmic function does not mention base value and it is well known the base is 10. hence, we can obtain the required result.
Complete step-by-step solution:
The logarithmic function is known as the inverse of exponential function. The logarithmic function is represented as \[{\log _b}a\], where b is a base number and a can be any numeral.
In the logarithmic functions we have two kinds namely,
Common logarithmic function: in this logarithmic function the base value is 10. It is represented as a log.
Natural logarithmic function: In this logarithmic function the base value is e (exponent). It is represented as ln.
Here in this question we have log, then it base value is 10.
Now consider the given function \[\log (x + 2) + \log (x - 1) = 1\]
By the property of logarithmic function we have \[\log a + \log b = \log (a.b)\], by using this property the given function is written as
\[ \Rightarrow \log ((x + 2)(x - 1)) = 1\]
The value of log 10 has 1. So the above equation is written as
\[ \Rightarrow \log ((x + 2)(x - 1)) = \log 10\]
Applying the anti log on the both sides o the above equation is written as
\[ \Rightarrow (x + 2)(x - 1) = 10\]
On multiplying the terms which are present in the LHS we have
\[ \Rightarrow x(x - 1) + 2(x - 1) = 10\]
\[
\Rightarrow {x^2} - x + 2x - 2 = 10 \\
\Rightarrow {x^2} + x - 12 = 0 \\
\]
The above equation in the form of quadratic equation. By factorising the above quadratic equation we get
\[ \Rightarrow {x^2} + 4x - 3x - 12 = 0\]
On simplifying we get
\[
\Rightarrow x(x + 4) - 3(x + 4) = 0 \\
\Rightarrow (x + 4)(x - 3) = 0 \\
\]
Therefore we have
\[ \Rightarrow x = - 4\] and \[x = 3\]
Hence we have solved the given logarithmic function
Note: The properties of logarithmic function does not vary for the common logarithmic function and natural logarithmic function. The value will change for the common logarithmic function and natural logarithmic function. We must know about the properties of logarithmic to solve these kinds of problems.
Complete step-by-step solution:
The logarithmic function is known as the inverse of exponential function. The logarithmic function is represented as \[{\log _b}a\], where b is a base number and a can be any numeral.
In the logarithmic functions we have two kinds namely,
Common logarithmic function: in this logarithmic function the base value is 10. It is represented as a log.
Natural logarithmic function: In this logarithmic function the base value is e (exponent). It is represented as ln.
Here in this question we have log, then it base value is 10.
Now consider the given function \[\log (x + 2) + \log (x - 1) = 1\]
By the property of logarithmic function we have \[\log a + \log b = \log (a.b)\], by using this property the given function is written as
\[ \Rightarrow \log ((x + 2)(x - 1)) = 1\]
The value of log 10 has 1. So the above equation is written as
\[ \Rightarrow \log ((x + 2)(x - 1)) = \log 10\]
Applying the anti log on the both sides o the above equation is written as
\[ \Rightarrow (x + 2)(x - 1) = 10\]
On multiplying the terms which are present in the LHS we have
\[ \Rightarrow x(x - 1) + 2(x - 1) = 10\]
\[
\Rightarrow {x^2} - x + 2x - 2 = 10 \\
\Rightarrow {x^2} + x - 12 = 0 \\
\]
The above equation in the form of quadratic equation. By factorising the above quadratic equation we get
\[ \Rightarrow {x^2} + 4x - 3x - 12 = 0\]
On simplifying we get
\[
\Rightarrow x(x + 4) - 3(x + 4) = 0 \\
\Rightarrow (x + 4)(x - 3) = 0 \\
\]
Therefore we have
\[ \Rightarrow x = - 4\] and \[x = 3\]
Hence we have solved the given logarithmic function
Note: The properties of logarithmic function does not vary for the common logarithmic function and natural logarithmic function. The value will change for the common logarithmic function and natural logarithmic function. We must know about the properties of logarithmic to solve these kinds of problems.
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