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How do you solve $\ln \left( 4x-2 \right)-\ln 4=-\ln \left( x-2 \right)$?

Last updated date: 23rd Jun 2024
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Hint:To solve the given question, first we apply the property of logarithm which states that if logs to the same base are added, then the numbers were multiplied, i.e. log (a) + log (b) = log (a.b). Then we simplify the equation further by using the definition of log, if log (a) = log (b) then a = b. and solve the equation in a way we solve the general quadratic equation.

Formula used:
The property of logarithm which states that if logs to the same base are added, then the
numbers were multiplied, i.e. log (a) + log (b) = log (a.b)
If log (a) = log (b) then a = b.

Complete step by step solution:
We have given that,
$\ln \left( 4x-2 \right)-\ln 4=-\ln \left( x-2 \right)$
Rearranging the terms in the above equation, we get
$\Rightarrow \ln \left( 4x-2 \right)+\ln \left( x-2 \right)=\ln 4$
Using the property of logarithm which states that if logs to the same base are added, then the numbers were multiplied, i.e. log (a) + log (b) = log (a.b)
Applying the above property, we get
$\Rightarrow \ln \left( \left( 4x-2 \right)\times \left( x-2 \right) \right)=\ln 4$
Using the definition of log, if log (a) = log (b) then a = b.
Applying the above property, we get
$\Rightarrow \left( \left( 4x-2 \right)\times \left( x-2 \right) \right)=\ln 4$
Simplifying the above equation, we get
$\Rightarrow \left( 4x\times x \right)+\left( 4x\times -2 \right)+\left( -2\times x \right)+\left( -2\times -2 \right)=4$
Simplifying further, we get
$\Rightarrow 4{{x}^{2}}-8x-2x+4=4$
$\Rightarrow 4{{x}^{2}}-10x+4=4$
Subtracting 4 from both the sides of the equation, we get
$\Rightarrow 4{{x}^{2}}-10x=0$
Taking out 2x as a common factor, we get
$\Rightarrow 2x\left( 2x-5 \right)=0$
Equation each factor equals to 0, we get
$\Rightarrow 2x=0$ and $2x-5=0$
Now, solving
$\Rightarrow 2x=0$
$\Rightarrow x=0$
Now, solving
$\Rightarrow 2x-5=0$
Adding 5 to both the sides of the equation, we get
$\Rightarrow 2x=5$
Dividing both the sides of the equation by 2, we get
$\Rightarrow x=\dfrac{5}{2}$
Since, x > 2, so the only possible value of x is $\dfrac{5}{2}$.
Therefore, $x=\dfrac{5}{2}$ is the required solution.

Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general quadratic equations.