
How do you solve ${\log _7}x + {\log _7}(x + 5) = {\log _7}(14)$?
Answer
558.6k+ views
Hint:In order to determine the value of the above question, rewrite the expression using the property of logarithm ${\log _b}(m) + {\log _b}(n) = {\log _b}(mn)$and take antilogarithm on both side to remove logarithm from the expression then use the splitting up the middle method to find the solution of the quadratic equation formed.
Formula:
${\log _b}(m) + {\log _b}(n) = {\log _b}(mn)$
Complete step by step solution:
We are Given an expression ${\log _7}x + {\log _7}(x + 5) = {\log _7}(14)$
Now, rewriting the expression using the property of logarithm ${\log _b}(m) + {\log _b}(n) = {\log
_b}(mn)$
${\log _7}\left( {x(x + 5)} \right) = {\log _7}14$
Taking antilogarithm on both sides ,this will remove the logarithm from both the sides, our
expression now becomes
\[
\Rightarrow x(x + 5) = 14 \\
\Rightarrow {x^2} + 5x = 14 \\
\Rightarrow {x^2} + 5x - 14 = 0 \\
\]
Expression has become a quadratic equation, and to solve this we’ll use splitting up the middle term method.
Follow below steps to split the middle term
Step 1: Calculate the product of coefficient of ${x^2}$and the constant term which comes to be $ = - 14 \times 1 = - 14$
Step 2: Find the 2 factors of the number -14 such that the weather addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant .
So if we factorize 14, the answer comes to be 7and 2 as \[7 - 2 = 5\] that is the middle term . and \[7 \times 2 = 14\] which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained, so equation becomes
\[
\Rightarrow {x^2} + 7x - 2x - 14 = 0 \\
\Rightarrow x(x + 7) - 2(x + 7) = 0 \\
\Rightarrow (x + 7)(x - 2) = 0 \\
\]
$
x + 7 = 0 \\
\Rightarrow x = - 7 \\
x - 2 = 0 \\
\Rightarrow x = 2 \\
$
Value of x can be $ - 7,2$
Since $\ln x$is not defined for the negative values of x so $x = 2$
Therefore, the solution to expression${\log _7}x + {\log _7}(x + 5) = {\log _7}(14)$is $x = 2$.
Note:
1.Value of constant ‘e’ is equal to $2.71828$.
2.A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number, we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values.
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values.
${\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$
6.Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become linear equation and will no more quadratic .
Formula:
${\log _b}(m) + {\log _b}(n) = {\log _b}(mn)$
Complete step by step solution:
We are Given an expression ${\log _7}x + {\log _7}(x + 5) = {\log _7}(14)$
Now, rewriting the expression using the property of logarithm ${\log _b}(m) + {\log _b}(n) = {\log
_b}(mn)$
${\log _7}\left( {x(x + 5)} \right) = {\log _7}14$
Taking antilogarithm on both sides ,this will remove the logarithm from both the sides, our
expression now becomes
\[
\Rightarrow x(x + 5) = 14 \\
\Rightarrow {x^2} + 5x = 14 \\
\Rightarrow {x^2} + 5x - 14 = 0 \\
\]
Expression has become a quadratic equation, and to solve this we’ll use splitting up the middle term method.
Follow below steps to split the middle term
Step 1: Calculate the product of coefficient of ${x^2}$and the constant term which comes to be $ = - 14 \times 1 = - 14$
Step 2: Find the 2 factors of the number -14 such that the weather addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant .
So if we factorize 14, the answer comes to be 7and 2 as \[7 - 2 = 5\] that is the middle term . and \[7 \times 2 = 14\] which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained, so equation becomes
\[
\Rightarrow {x^2} + 7x - 2x - 14 = 0 \\
\Rightarrow x(x + 7) - 2(x + 7) = 0 \\
\Rightarrow (x + 7)(x - 2) = 0 \\
\]
$
x + 7 = 0 \\
\Rightarrow x = - 7 \\
x - 2 = 0 \\
\Rightarrow x = 2 \\
$
Value of x can be $ - 7,2$
Since $\ln x$is not defined for the negative values of x so $x = 2$
Therefore, the solution to expression${\log _7}x + {\log _7}(x + 5) = {\log _7}(14)$is $x = 2$.
Note:
1.Value of constant ‘e’ is equal to $2.71828$.
2.A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number, we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values.
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values.
${\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$
6.Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become linear equation and will no more quadratic .
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

