Question

How do you solve the equation ${{\log }_{2}}\left( \dfrac{3x-2}{5} \right)={{\log }_{2}}\left( \dfrac{8}{x} \right)$ ?

Answer 1

Hint: To solve this question first check whether the base of both the logarithmic expressions are the same, if they are the same then remove the log from both sides and solve the equation for the values of the variable. After finding the values of the variable, substitute them in the equation to check which value satisfies the expression to get the final answer.

Complete step by step solution:
Given the expression: ${{\log }_{2}}\left( \dfrac{3x-2}{5} \right)={{\log }_{2}}\left( \dfrac{8}{x} \right)$
Since, the base on both the sides is the same, that is $2$ , therefore we can easily remove the log from both the sides, to get the following equation,
$\Rightarrow \left( \dfrac{3x-2}{5} \right)=\left( \dfrac{8}{x} \right)$
Now, simplify the above expression by cross-multiplying the denominators of both the fractions, to get,
 $\Rightarrow x\left( 3x-2 \right)=5\times 8$
Simplify the above expression by multiplying the terms,
$\Rightarrow 3{{x}^{2}}-2x=40$
Rewrite the above equation the standard form of a quadratic equation, to get,
$\Rightarrow 3{{x}^{2}}-2x-40=0$
Now, with the help of splitting the middle term method, solve the above equation to get the value of the variable $x$ ,
$\Rightarrow 3{{x}^{2}}-12x+10x-40=0$
The middle term $2x\;$ can be split like $-12x$ and $10x\;$, since the product of the first and the last term is $-120$ .
Taking the common terms from the above equation and writing it in the form of products of factors, we get,
 $\Rightarrow 3x\left( x-4 \right)+10\left( x-4 \right)=0$
Taking out the common term,
$\Rightarrow \left( 3x+10 \right)\left( x-4 \right)=0$
From the above equation and by equating each term to zero, we get the following values for the variable $x$
$x=\dfrac{-10}{3}$ Or $x=4$
Now, we know that logarithmic functions are only defined on the positive real number set, that is, the domain of the logarithmic functions is real positive numbers only.
So, we can say that $x=\dfrac{-10}{3}$ would not be a valid solution.
Therefore, on solving ${{\log }_{2}}\left( \dfrac{3x-2}{5} \right)={{\log }_{2}}\left( \dfrac{8}{x} \right)$ we get the value as $x=4$.

Note: While solving the quadratic equations, one can use the quadratic formula as well as splitting the middle term as well. Also, keep in mind that the exponential and logarithmic functions are only defined on the set of positive real numbers.