Find the $$ x \quad\left(\frac{4}{9} ight)^{4} imes\left(\frac{4}{9} ight)^{-7}=\frac{4}{9}{ }^{2 x-1} $$

Question

Find the $$ x \quad\left(\frac{4}{9}ight)^{4} 	imes\left(\frac{4}{9}ight)^{-7}=\frac{4}{9}{ }^{2 x-1} $$

UpStudy AI · Accepted Answer

To solve the equation:

$$
\left(\frac{4}{9}\right)^{4} \times \left(\frac{4}{9}\right)^{-7} = \left(\frac{4}{9}\right)^{2x - 1}
$$

follow these steps:

1. **Combine the Exponents on the Left Side:**

Use the **product of powers** property, which states that $$ a^m \times a^n = a^{m+n} $$.

$$
   \left(\frac{4}{9}\right)^{4} \times \left(\frac{4}{9}\right)^{-7} = \left(\frac{4}{9}\right)^{4 - 7} = \left(\frac{4}{9}\right)^{-3}
   $$

2. **Set the Exponents Equal:**

Since the bases are the same, set the exponents equal to each other.

$$
   -3 = 2x - 1
   $$

3. **Solve for $$ x $$:**

$$
   -3 + 1 = 2x \
   -2 = 2x \
   x = \frac{-2}{2} \
   x = -1
   $$

**Answer:** $$ x = -1 $$
$$ x = -1 $$

Find the \( x \quad\left(\frac{4}{9}\right)^{4} \times\left(\frac{4}{9}\right)^{-7}=\frac{4}{9}{ }^{2 x-1} \)