Math Doubts

Find the value of $m-n$, if $\dfrac{9^n \times 3^2 \times 3^n -27^n}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

$\dfrac{9^n \times 3^2 \times 3^n -27^n}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

$27^n$ is an exponential term and it is subtracted from the product of the exponential terms $9^n$, $3^2$ and $3^n$ and the entire subtraction is divided by the product of $3^{3m}$ and $2^3$. The quotient of them is equal to the $\dfrac{1}{27}$. The value of $m-n$ is required to find by simplifying the equation.

Step: 1

Convert the exponential terms in the numerator to make every term contains number $3$ as base of the exponential term. It can be done by using power rule of exponents.

$\implies \dfrac{{(3^2)}^n \times 3^2 \times 3^n -{(3^3)}^n}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

Step: 2

The second term is $3^{3n}$ in numerator. The same quantity can be obtained by combining exponential terms $3^{2n}$ and $3^n$.

$\implies \dfrac{3^{2n} \times 3^2 \times 3^n -3^{3n}}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

$\implies \dfrac{3^{2n} \times 3^n \times 3^2 -3^{3n}}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

Apply product rule of exponents to combine $3^{2n}$ and $3^n$ as a single exponential term $3^{3n}$.

$\implies \dfrac{3^{2n+n} \times 3^2 -3^{3n}}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

$\implies \dfrac{3^{3n} \times 3^2 -3^{3n}}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

Step: 3

The exponential term $3^{3n}$ is a common term in the expression of the numerator. So, take it common from both terms.

$\implies \dfrac{3^{3n}(3^2 -1)}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

$\implies \dfrac{3^{3n}(9 -1)}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

$\implies \dfrac{3^{3n}(8)}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

The number $8$ can be expressed as $2^3$ due to the same quantity in the denominator.

$\implies \dfrac{3^{3n} \times 2^3}{3^{3m} \times 2^3}$ $=$ $\dfrac{1}{27}$

Step: 4

Cancel the $2^3$ by the same once to simplify it further.

$\implies \require{cancel} \dfrac{3^{3n} \times \cancel{2^3}}{3^{3m} \times \cancel{2^3}}$ $=$ $\dfrac{1}{27}$

$\implies \dfrac{3^{3n}}{3^{3m}}$ $=$ $\dfrac{1}{27}$

Step: 5

The value of $m-n$ is required to find. So, move $3^{3n}$ to denominator to combine both terms.

$\implies \dfrac{1}{3^{3m} \times 3^{-3n}}$ $=$ $\dfrac{1}{27}$

Apply product law of exponents to combine them as a single exponential term.

$\implies \dfrac{1}{3^{3m-3n}}$ $=$ $\dfrac{1}{27}$

$\implies \dfrac{1}{3^{3m-3n}}$ $=$ $\dfrac{1}{3^3}$

Step: 6

Use cross multiplication method to simplify the equation further.

$\implies 3^{3m-3n} = 3^3$

$\implies 3m-3n = 3$

$\implies 3(m-n) = 3$

$\implies m-n = \dfrac{3}{3}$

$\therefore \,\,\,\,\,\, m-n = 1$

The value of $m-n$ is $1$ and it is the required solution for this problem in mathematics.



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