Question

Solution of $\log \dfrac{5}{4}+\log 14-\log \dfrac{7x}{3}=-1$ is

Answer 1

Hint: To answer this question, we need to know the basic properties of log function. We need to use the product and quotient rules of logarithms in order to group all the terms on the left-hand side into one term. Then we take the antilog on both sides of the equation and simplify for the variable x.

Complete step by step solution:
The given question requires us to know a few important log properties. First, we define the few important rules that we are going to use in order to solve this question. We are going to use what is known as the product rule here. Product rule says that log of a product of two terms can be written as the sum of log of the two terms.
$\log a.b=\log a+\log b\ldots \ldots \left( 1 \right)$
We also use the quotient rule which says that the log of the division of two terms can be written as the log of the numerator term subtracted by the log of the denominator term.
$\log \dfrac{a}{b}=\log a-\log b\ldots \ldots \left( 2 \right)$
Now, we solve the left-hand side of this equation.
$\Rightarrow \log \dfrac{5}{4}+\log 14-\log \dfrac{7x}{3}$
We apply the product rule given by equation $\left( 1 \right)$ for the first two terms and simplify it by writing it as a single log term.
$\Rightarrow \log \dfrac{5}{4}\times 14-\log \dfrac{7x}{3}$
We apply the quotient rule given by equation $\left( 2 \right)$ to the above two terms and simplify it by writing it as a single log term with the division of the first term by the second term.
$\Rightarrow \log \dfrac{\dfrac{5}{4}\times 14}{\dfrac{7x}{3}}$
We know that if two terms are divided, then it is equivalent to the product of the reciprocated term.
$\Rightarrow \log \dfrac{5}{4}\times 14\times \dfrac{3}{7x}$
The numerator has a term 14 and the denominator has the term 7 both of which have a HCF of 7 and hence can be simplified as,
$\Rightarrow \log \dfrac{5}{4}\times 2\times \dfrac{3}{x}$
Similarly, the HCF of the terms 2 in the numerator and 4 in the denominator is 2 and cancelling these,
$\Rightarrow \log \dfrac{5}{2}\times 1\times \dfrac{3}{x}$
We can see that there are no more factors left to simplify this further. Multiplying all the numerators and denominators separately and writing it down,
$\Rightarrow \log \dfrac{15}{2x}$
Now we equate this to the right-hand side,
$\Rightarrow \log \dfrac{15}{2x}=-1$
Taking anit-log on both sides,
$\Rightarrow anti\log \left( \log \dfrac{15}{2x} \right)=anti\log \left( -1 \right)$
The log and anti-log are inverses of each other and cancel on the left-hand side.
$\Rightarrow \dfrac{15}{2x}=anti\log \left( -1 \right)\ldots \ldots \left( 3 \right)$
We also know that the antilog can be written as follows,
If ${{\log }_{b}}a=x,$ then $a=anti\log \left( x \right).$
Therefore, we use this for the above equation considering x as -1.
$\Rightarrow a=anti\log \left( -1 \right)$
Here the base is 10.
$\Rightarrow {{\log }_{10}}a=-1$
We know the log definition that if ${{\log }_{b}}a=x,$ then ${{b}^{x}}=a.$
This means that,
$\Rightarrow {{10}^{-1}}=a$
The left-hand side is nothing but 1 divided by 10.
$\Rightarrow \dfrac{1}{10}=a$
Hence the right-hand side of equation $\left( 3 \right)$ is equal to $\dfrac{1}{10}.$ Substituting this is the equation,
$\Rightarrow \dfrac{15}{2x}=\dfrac{1}{10}$
Cross multiplying both sides of the equation,
$\Rightarrow 15\times 10=1\times 2x$
Taking the product of 15 and 10 and dividing both sides by x,
$\Rightarrow \dfrac{150}{2}=\dfrac{2}{2}x$
Simplifying this by dividing 150 by 2,
$\Rightarrow 75=x$
Hence, the solution to the above equation is x=75.

Note: Students need to know the basic and important rules of logarithms in order to solve this question. It is important to note that the base for all the log terms here in this question are 10. If not specified, we need to consider the base as 10 for any problem containing log terms. And, if not specified, we need to consider the base as e for any problem containing ln terms.