Question

How do you solve for $x$ in ${{\log }_{6}}\left( 2-x \right)+{{\log }_{6}}\left( 3-4x \right)=1$?

Answer 1

Hint: In this question we have been given with a logarithmic expression we will use the property of logarithm $\log \left( a \right)+\log \left( b \right)=\log \left( ab \right)$ and then multiply the terms. we will then remove the log on both sides by using the property of logarithm ${{\log }_{a}}y=x\to {{a}^{x}}=y$. We will then find the values of $x$ and check by substituting the values in the equation whether they are the true solution or an extraneous solution.

Complete step-by-step answer:
We have the expression given to us as:
$\Rightarrow {{\log }_{6}}\left( 2-x \right)+{{\log }_{6}}\left( 3-4x \right)=1\to \left( 1 \right)$
We can see that the expression is in the form of addition of logarithm therefore, we will use the property $\log \left( a \right)+\log \left( b \right)=\log \left( ab \right)$ to combine them.
On using the property, we get:
$\Rightarrow {{\log }_{6}}\left( \left( 2-x \right)\left( 3-4x \right) \right)=1$
On using the property of logarithm ${{\log }_{a}}y=x\to {{a}^{x}}=y$ on the expression, we get:
$\Rightarrow \left( 2-x \right)\left( 3-4x \right)={{6}^{1}}$
On multiplying the terms, we get:
$\Rightarrow 6-8x-3x+4{{x}^{2}}=6$
On simplifying and rearranging the terms, we get:
$\Rightarrow 4{{x}^{2}}-11x+6=6$
On subtracting the term $6$ from both the sides of the expression, we get:
$\Rightarrow 4{{x}^{2}}-11x=0$
On taking the term $x$ common from both the sides, we get:
$\Rightarrow x\left( 4x-11 \right)=0$
Now we know the property that when $ab=0$, $a=0$ or $b=0$.
Therefore, we get:
$x=0$ and $4x-11=0$
On rearranging the terms, we get:
$x=0$ and $x=\dfrac{11}{4}$, which are the required solutions.
On substituting $x=\dfrac{11}{4}$ in equation $\left( 1 \right)$, we get:
$\Rightarrow {{\log }_{6}}\left( 2-\dfrac{11}{4} \right)+{{\log }_{6}}\left( 3-4\left( \dfrac{11}{4} \right) \right)=1$
On simplifying the terms, we get:
$\Rightarrow {{\log }_{6}}\left( -\dfrac{3}{4} \right)+{{\log }_{6}}\left( -8 \right)=1$
Now since logarithm of negative numbers does not exist, $x=\dfrac{11}{4}$ is an extraneous solution.
On substituting $x=0$ in equation $\left( 1 \right)$, we get:
$\Rightarrow {{\log }_{6}}\left( 2-0 \right)+{{\log }_{6}}\left( 3-4\left( 0 \right) \right)=1$
On simplifying, we get:
$\Rightarrow {{\log }_{6}}2+{{\log }_{6}}3=1$
On using the property $\log \left( a \right)+\log \left( b \right)=\log \left( ab \right)$, we get:
$\Rightarrow {{\log }_{6}}6=1$
since we know that ${{\log }_{a}}a=1$, we get:
$\Rightarrow 1=1$, since the left-hand side is equal to the right-hand side $x=0$ is the correct solution.

Note: It is to be noted that the logarithm we are using has the base $6$, the base is the number to which the log value has to be raised to, to get the original term. This is also called the antilog of the number which is the logical reverse of taking a log.
The most commonly used bases in logarithm are $10$ and $e$ which has a value of approximate $2.713...$
Logarithm is used to simplify a mathematical expression, it converts multiplication to addition, division to subtraction and exponents to multiplication.