Question

How do you solve the exponential equation ${{9}^{2x+1}}={{3}^{5x-1}}$?

Answer 1

We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities We find the relation between negative exponent and inverse of the number to find the solution.

Complete step-by-step solution:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
For our given equation ${{9}^{2x+1}}={{3}^{5x-1}}$, we convert all the given numbers as the power of value 3. We know that $9={{3}^{2}}$.
If we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
We also have the identity of ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
Therefore, for the left-hand side of the equation ${{9}^{2x+1}}={{\left( {{3}^{2}} \right)}^{2x+1}}={{3}^{4x+2}}$.
We have the final equation where ${{3}^{4x+2}}={{3}^{5x-1}}$.
Now we know that if the bases are equal and power are different as ${{a}^{m}}={{a}^{n}}$ then $m=n$.
For the equation ${{3}^{4x+2}}={{3}^{5x-1}}$, we get $4x+2=5x-1$ which gives
$\begin{aligned} & 4x+2=5x-1 \\\ & \Rightarrow 5x-4x=2+1 \\\ & \Rightarrow x=3 \\\ \end{aligned}$.
Therefore, solving the equation ${{9}^{2x+1}}={{3}^{5x-1}}$ we get $x=3$.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$. We need to remember that the condition for ${{a}^{m}}={{a}^{n}}\Rightarrow m=n$ is that the value of $a\ne 0,\pm 1$.