-3) $$ \frac{3^{2}-3^{x-1}}{5 \cdot 3^{x}-3^{x+1}} $$

Question

UpStudy AI · Accepted Answer

We start with the expression

$$
\frac{3^2-3^{x-1}}{5 \cdot 3^x-3^{x+1}}.
$$

**Step 1. Simplify the numerator**

Note that

$$
3^2=9.
$$

So the numerator becomes

$$
9-3^{x-1}.
$$

We can write $$3^{x-1}$$ as

$$
3^{x-1}=\frac{3^x}{3}.
$$

Then the numerator is

$$
9-\frac{3^x}{3}=\frac{27-3^x}{3},
$$

since $$9=\frac{27}{3}$$.

**Step 2. Simplify the denominator**

The denominator is

$$
5\cdot 3^x-3^{x+1}.
$$

Recognize that

$$
3^{x+1}=3^x\cdot 3.
$$

Thus, the denominator becomes

$$
5\cdot 3^x-3^x\cdot 3=3^x(5-3)=2\cdot 3^x.
$$

**Step 3. Write the combined fraction and simplify**

Substitute back the simplified numerator and denominator:

$$
\frac{\frac{27-3^x}{3}}{2\cdot 3^x}=\frac{27-3^x}{3\cdot 2\cdot 3^x}=\frac{27-3^x}{6\cdot 3^x}.
$$

**Step 4. Factor the numerator**

Notice that $$27=3^3$$, hence the numerator is

$$
3^3-3^x.
$$

Factor $$3^x$$ from the numerator:

$$
3^3-3^x=3^x\left(3^{3-x}-1\right).
$$

Thus, the expression becomes

$$
\frac{3^x\left(3^{3-x}-1\right)}{6\cdot 3^x}.
$$

Cancel the common $$3^x$$ (which is always nonzero):

$$
\frac{3^{3-x}-1}{6}.
$$

**Final Answer**

$$
\frac{3^{2}-3^{x-1}}{5 \cdot 3^{x}-3^{x+1}}=\frac{3^{3-x}-1}{6}.
$$
$$ \frac{3^{3-x}-1}{6} $$

-3) \( \frac{3^{2}-3^{x-1}}{5 \cdot 3^{x}-3^{x+1}} \)