Question

Given $\log 7 - \log 2 + \log 16 - 2\log 3 - \log \dfrac{7}{{45}} = 1 + \log n,{\text{find }}n$

Answer 1

Hint: Here we have to find the value of $n$ by using the properties of logarithm. On doing some simplification we get the required answer. We will make use of the below mentioned identities to get the answer.
$\log \dfrac{a}{b} = \log a - \log b \to (1)$
$\log {a^b} = b\log a \to (2)$
$\log a \times b = \log a + \log b \to (3)$

Complete step-by-step solution:
It is given that the equation: $\log 7 - \log 2 + \log 16 - 2\log 3 - \log \dfrac{7}{{45}} = 1 + \log n$
First we take the left-hand side of the equation we get:
$ \Rightarrow \log 7 - \log 2 + \log 16 - 2\log 3 - \log \dfrac{7}{{45}}$
Now we know that the number $16$ can be expressed in terms of the power of $2$ as ${2^4}$ therefore, on re-writing the equation we get:
$ \Rightarrow \log 7 - \log 2 + \log {2^4} - 2\log 3 - \log \dfrac{7}{{45}}$
We also know the property of division of logarithm which is: $\log \dfrac{a}{b} = \log a - \log b \to (1)$
Therefore, on re-writing the equation we get:
$ \Rightarrow \log 7 - \log 2 + \log {2^4} - 2\log 3 - (\log 7 - \log 45)$
On opening the bracket, we get:
$ \Rightarrow \log 7 - \log 2 + \log {2^4} - 2\log 3 - \log 7 + \log 45$
On simplifying we get:
$ \Rightarrow - \log 2 + \log {2^4} - 2\log 3 + \log 45$
Now we know the property of logarithm which is: $\log {a^b} = b\log a \to (2)$therefore, on re-writing the equation we get:
$ \Rightarrow - \log 2 + 4\log 2 - 2\log 3 + \log 45$
On simplifying the equation, we get:
$ \Rightarrow 3\log 2 - 2\log 3 + \log 45$
Now using the property $(2)$we can re-write the equation as:
$ \Rightarrow 3\log 2 - \log {3^2} + \log 45$
It can be squared and written as:
$ \Rightarrow 3\log 2 - \log 9 + \log 45$
Now since the number $45$ can be written as a multiplication of $9 \times 5$ therefore, on re-writing the equation we get:
$ \Rightarrow 3\log 2 - \log 9 + \log (9 \times 5)$
Now we know the property of logarithm which is: $\log a \times b = \log a + \log b \to (3)$ therefore, on re-writing the equation we get:
$ \Rightarrow 3\log 2 - \log 9 + \log 9 + \log 5$
On simplifying we get:
$ \Rightarrow 3\log 2 + \log 5$
Now on splitting the term $3\log 2$as $2\log 2 + \log 2$ we can write the equation as:
$ \Rightarrow 2\log 2 + \log 2 + \log 5$
Now on using the property $(3)$ we get:
$ \Rightarrow 2\log 2 + \log (2 \times 5)$
On simplifying we get:
$ \Rightarrow 2\log 2 + \log 10$
Now we know that $\log 10 = 1$therefore we get:
$ \Rightarrow 2\log 2 + 1$
Now using the property $(2)$we get:
$ \Rightarrow \log {2^2} + 1$
On simplifying we get:
$ \Rightarrow \log 4 + 1$, here we get the value of LHS.
Now right-hand side is: $1 + \log n$
Therefore, the original equation can be written as:
$\log 4 + 1 = 1 + \log n$
On rephrasing the equation, we get:
$1 + \log 4 = 1 + \log n$

Therefore, using the property of equality, we know $n = 4$

Note: Logarithm coined as log is used for the multiplication and division of large numbers by converting the values of the numbers into addition and then removing the antilog of them.
Every log number has a base value $n$, the base is the number to which the value of the log has to be raised to get the original number.
The most commonly used bases in log are $10$ and $e$ which have a value $2.713...$which is an irrational number.