Question

Solve $ {{\log }_{5-x}}\left( {{x}^{2}}-2x+65 \right)=2 $ and find the value of x.

Answer 1

Hint: In this question, we are given the expression as $ {{\log }_{5-x}}\left( {{x}^{2}}-2x+65 \right)=2 $ and we need to find the value of x. For this, we will first use the property of logarithm that is $ {{\log }_{b}}x=y\Rightarrow x={{b}^{y}} $ . Then we will get an equation in terms of x. Solving it will give us the value of x. We will use the arithmetic property of the square of the difference between two terms which is $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ .

Complete step by step answer:
Here we are given the equation as $ {{\log }_{5-x}}\left( {{x}^{2}}-2x+65 \right)=2 $ .
We need to solve it to find the value of x. For this, we will use the property of logarithm according to which logarithmic equation can be expressed in exponential form using the relation: $ {{\log }_{b}}x=y\Rightarrow x={{b}^{y}} $ .
Here we have b = 5-x, $ x={{x}^{2}}-2x+65 $ and y = 2.
Solving, $ {{\log }_{5-x}}\left( {{x}^{2}}-2x+65 \right)=2 $ in same way we get, \[\left( {{x}^{2}}-2x+65 \right)={{\left( 5-x \right)}^{2}}\].
This is our equation that we need to solve. Now, we know from the arithmetic property of the square of the difference between two terms is given by $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . So let us use it on the right side of the equation assuming a as 5 and b as x, we get, \[{{x}^{2}}-2x+65={{\left( 5 \right)}^{2}}+{{\left( x \right)}^{2}}-2\left( 5 \right)\left( x \right)\].
We know that $ {{5}^{2}} $ can be written as $ 5\times 5 $ which is equal to 25, so we get, \[{{x}^{2}}-2x+65=25+{{x}^{2}}-10x\].
Taking a variable to one side and constant to the other side we get, \[{{x}^{2}}-2x-{{x}^{2}}+10x=25-65\].
Cancelling \[{{x}^{2}}\] by \[{{x}^{2}}\] and solving the rest of the terms we get, $ 8x=-40 $ .
Dividing both sides by 8 we get, $ x=\dfrac{-40}{8}=-5 $ .
Hence the value of x is -5 which is our required solution.

Note:
 Students should take care of the signs while solving this equation. Keep in mind the properties of arithmetic to solve this equation. Students can also check their answers by following way,
Putting x = -5 in the given equation we get,
 $ {{\log }_{5-x}}\left( {{\left( -5 \right)}^{2}}-2\left( -5 \right)+65 \right)=2 $ .
Simplifying we get, $ {{\log }_{5-x}}\left( 25+10+65 \right)=2\Rightarrow {{\log }_{5-x}}\left( 100 \right)=2 $ .
Now 100 can be written as $ {{10}^{2}} $ so we get,
 $ {{\log }_{5-x}}{{\left( 10 \right)}^{2}}=2 $ .
By properties of logarithm, $ {{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}\text{ and }{{\log }_{b}}b=1 $ we get,
 $ 2{{\log }_{10}}10=2\Rightarrow 2\left( 1 \right)=2\Rightarrow 2=2 $ .
Hence our answer is correct.