Question

How do you solve $\log 6x-3=-4$?

Answer 1

In the given question, we have been asked to solve $\log 6x-3=-4$ and find the value of ‘x’. In order to solve the given question, first we need to take all the numbers on one side of the equation. Later using the definition of log which states that if $x$ and b are positive real numbers and b is not equal to 1, then ${{\log }_{b}}\left( x \right)=y$ is equivalent to ${{b}^{y}}=x$. And then we solve the equation in a way we solve general linear equations.

Complete step by step solution:
We have given,
$\log 6x-3=-4$
Transposing -3 to the right side of the equation it will become +3, we get
$\Rightarrow \log 6x=-4+3$
Adding the numbers on the right side of the equation, we get
$\Rightarrow \log 6x=-1$
Now,
Using the definition of log,
If $x$ and b are positive real numbers and b is not equal to 1,
Then ${{\log }_{b}}\left( x \right)=y$is equivalent to${{b}^{y}}=x$.
Therefore,
$\log \left( 6x \right)=\log \left( 0.1 \right)$
Using the definition of log, if log (a) = log (b) then a = b.
Thus,
$\Rightarrow 6x=0.1$
Now, solving for the value of ‘x’, we get
$\Rightarrow x=\dfrac{0.1}{6}$
After removing the decimal, we get
$\Rightarrow x=\dfrac{1}{60}$
Therefore, the value of $x=\dfrac{1}{60}$ is the required solution.
Formula used:
If $x$ and b are positive real numbers and b is not equal to 1,
Then ${{\log }_{b}}\left( x \right)=y$ is equivalent to ${{b}^{y}}=x$.
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).

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.