# Solve $9 \times 81^{\large x} = \dfrac{1}{27^{\large x \normalsize-3}}$

The given exponential equation is $9 \times 81^{\large x} = \dfrac{1}{27^{\large x \normalsize-3}}$, where $x$ is a variable and represents a real number.

In this exponential equation, $9$, $27$ and $81$ are multiplies of number $3$. So, they can be written in exponential form in terms of $3$ by exponentiation.

$\implies$ $3^2 \times \Big(3^4\Big)^{\large x} = \dfrac{1}{\Big(3^3\Big)^{\large x \normalsize-3}}$

Now, use power rule of exponents to simplify the factors which contains powers of exponential form quantities.

$\implies$ $3^2 \times 3^{4 \times \large x} = \dfrac{1}{3^{3 \times (\large x \normalsize-3)}}$

$\implies$ $3^2 \times 3^{4\large x} = \dfrac{1}{3^{3(\large x \normalsize-3)}}$

The given exponential equation is successfully simplified and it has to be solved for evaluating the $x$. It can be done in two different methods.

### Beginner’s Method

In the left hand side of the exponential equation, the factors have same bases. The term can be simplified by using the product rule of exponents with same base.

$\implies$ $3^{2+4\large x} = \dfrac{1}{3^{3(\large x \normalsize-3)}}$

In both sides of the exponential equation, the exponential terms contain same base. So, they can be merged if the two exponential terms are shifted to one side of the equation. It can be done by the cross multiplication technique.

$\implies$ $3^{2+4\large x} \times 3^{3(\large x \normalsize-3)} = 1$

Now, use product rule of exponents with same base to merge them as an exponential term.

$\implies$ $3^{2+4\large x \normalsize+ 3(\large x \normalsize-3)} = 1$

Look at the exponential equation. The left hand side expression contains $3$ as base but the right hand side is $1$. It is not possible to solve them. A special technique has to use to move ahead in solving this equation. According to zero power rule, the power $0$ with base $3$ is equal to one. So, the number $1$ can be written as $3^0$.

$\implies$ $3^{2+4\large x \normalsize+ 3(\large x \normalsize-3)} = 3^0$

The bases of both sides of the equation are same. So, the exponents should be equal logically.

$\implies$ $2+4x+3(x-3) = 0$

Now, simplify this algebraic equation to calculate the variable $x$.

$\implies$ $2+4x+3x-9 = 0$

$\implies$ $4x+3x-9+2 = 0$

$\implies$ $7x-7 = 0$

It is a linear equation in one variable, solve it in any one of the methods of solving linear equations in one variable.

$\implies$ $7x = 7$

$\implies$ $x = \dfrac{7}{7}$

$\implies$ $\require{cancel} x = \dfrac{\cancel{7}}{\cancel{7}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 1$

Therefore, it is solved that $x$ is equal to $1$.

The given exponential equation $9 \times 81^{\large x} = \dfrac{1}{27^{\large x \normalsize-3}}$ is simplified above in the following form.

$\implies$ $3^{2+4\large x} = \dfrac{1}{3^{3(\large x \normalsize-3)}}$

It can be simply solved directly by using negative power rule of exponents.

$\implies$ $3^{2+4\large x} = 3^{-3(\large x \normalsize-3)}$

The bases of exponential terms in both sides of the equation are same. So, the exponents should be equal mathematically.

$\implies$ $2+4x = -3(x-3)$

Now, simplify this algebraic equation for solving the given exponential equation.

$\implies$ $2+4x = -3x+9$

$\implies$ $4x+3x = 9-2$

It is a linear equation in one variable and solve the $x$.

$\implies$ $7x = 7$

$\implies$ $x = \dfrac{7}{7}$

$\implies$ $\require{cancel} x = \dfrac{\cancel{7}}{\cancel{7}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 1$

In this way, the given exponential equation $9 \times 81^{\large x} = \dfrac{1}{27^{\large x \normalsize-3}}$ can be solved mathematically in two different methods.

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