If $$ f(x)=(5)^{-x} $$, then $$ \frac{f(x-1)}{f(x+1)}=\ldots \ldots $$

Question

UpStudy AI · Accepted Answer

To solve for $$ \frac{f(x-1)}{f(x+1)} $$ where $$ f(x) = (5)^{-x} $$, we first need to express $$ f(x-1) $$ and $$ f(x+1) $$.

1. Calculate $$ f(x-1) $$:
   $$
   f(x-1) = (5)^{-(x-1)} = (5)^{-x + 1} = (5)^{-x} \cdot (5)^{1} = 5 \cdot (5)^{-x}
   $$

2. Calculate $$ f(x+1) $$:
   $$
   f(x+1) = (5)^{-(x+1)} = (5)^{-x - 1} = (5)^{-x} \cdot (5)^{-1} = (5)^{-x} \cdot \frac{1}{5}
   $$

Now we can substitute these expressions into the fraction:

$$
\frac{f(x-1)}{f(x+1)} = \frac{5 \cdot (5)^{-x}}{(5)^{-x} \cdot \frac{1}{5}}
$$

Next, we simplify the fraction:

$$
= \frac{5 \cdot (5)^{-x}}{(5)^{-x} \cdot \frac{1}{5}} = \frac{5}{\frac{1}{5}} = 5 \cdot 5 = 25
$$

Thus, we find that:

$$
\frac{f(x-1)}{f(x+1)} = 25
$$

Therefore, the final answer is:

$$
\boxed{25}
$$
$$\frac{f(x-1)}{f(x+1)} = 25$$

If \( f(x)=(5)^{-x} \), then \( \frac{f(x-1)}{f(x+1)}=\ldots \ldots \)