Question 2 (1 point) Identify the common base that should be used to solve the equation below. $$ 9^{5 x}=27^{2 x-1} $$ 27 $$ \frac{18}{3} $$

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To solve the equation $$ 9^{5x} = 27^{2x - 1} $$, we need to express both sides of the equation with a common base.

1. **Identify the bases:**
   - The left side has a base of $$ 9 $$.
   - The right side has a base of $$ 27 $$.

2. **Express both bases in terms of a common base:**
   - We can express $$ 9 $$ as $$ 3^2 $$ because $$ 9 = 3^2 $$.
   - We can express $$ 27 $$ as $$ 3^3 $$ because $$ 27 = 3^3 $$.

3. **Rewrite the equation using the common base $$ 3 $$:**
   - The left side becomes $$ (3^2)^{5x} = 3^{10x} $$.
   - The right side becomes $$ (3^3)^{2x - 1} = 3^{3(2x - 1)} = 3^{6x - 3} $$.

4. **Set the exponents equal to each other:**
   $$
   10x = 6x - 3
   $$

Now we can solve for $$ x $$.

Let's proceed with the calculation.

1. Subtract $$ 6x $$ from both sides:
   $$
   10x - 6x = -3
   $$
   $$
   4x = -3
   $$

2. Divide by $$ 4 $$:
   $$
   x = -\frac{3}{4}
   $$

Thus, the common base used to solve the equation is $$ 3 $$.
The common base used to solve the equation is 3.

Question 2 (1 point) Identify the common base that should be used to solve the equation below. \( 9^{5 x}=27^{2 x-1} \) 27 \( \frac{18}{3} \)

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