Question

How do you solve$\log \left( x-1 \right)+\log 2=\log \left( 3x \right)$?

Answer 1

In the given question, we have been asked to find the value of ‘x’ and it is given that$\log \left( x-1 \right)+\log 2=\log \left( 3x \right)$. In order to find the value of ‘x’, first we will apply the law of logarithm which states that$\log a+\log b=\log \left( a\times b \right)$. Later using the distributive property of multiplication i.e. $a\times \left( b+c \right)=\left( a\times b \right)+\left( a\times c \right)$. We will simplify the question. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations.

Complete step by step solution:
We have given,
$\log \left( x-1 \right)+\log 2=\log \left( 3x \right)$
By using the law of logarithm, i.e.
$\log a+\log b=\log \left( a\times b \right)$
Applying the law of logarithm, we get
$\Rightarrow \log \left( \left( x-1 \right)\times 2 \right)=\log \left( 3x \right)$
Using the distributive property, i.e.
$a\times \left( b+c \right)=\left( a\times b \right)+\left( a\times c \right)$
Applying this property, we obtain
$\Rightarrow \log \left( 2x-2 \right)=\log \left( 3x \right)$
We can write it as;
$\Rightarrow {{10}^{\log \left( 2x-2 \right)}}={{10}^{\log \left( 3x \right)}}$
Therefore,
$\Rightarrow 2x-2=3x$
Now, solving for the value of ‘x’.
Subtracting 2 from both the sides of the equation, we get
$\Rightarrow 2x-2-2x=3x-2x$
Simplifying the above equation, we get
$\Rightarrow -2=x$
Therefore, $x=-2$
Since you cannot apply log function on a negative number, there is no solution for this logarithmic equation.
There is no value of ‘x’ that makes the equation true because log cannot be applied on a negative number.
Formula used:
The law of logarithm states that $\log a+\log b=\log \left( a\times b \right)$
The distributive property of multiplication, i.e. $a\times \left( b+c \right)=\left( a\times b \right)+\left( a\times c \right)$

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 be required 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 linear equations. We should have remembered that the argument of log number can only take positive argument i.e. positive numbers.