# Leetcode - Power of Three Solution

Given an integer `n`

, return * true if it is a power of three. Otherwise, return false*.

An integer `n`

is a power of three, if there exists an integer `x`

such that `n == 3`

.^{x}

**Example 1:**

```
Input: n = 27
Output: true
```

**Example 2:**

```
Input: n = 0
Output: false
```

**Example 3:**

```
Input: n = 9
Output: true
```

**Example 4:**

```
Input: n = 45
Output: false
```

**Constraints:**

`-2`

^{31}<= n <= 2^{31}- 1

**Follow up:** Could you solve it without loops/recursion?

## Solution in python

```
class Solution:
def isPowerOfThree(self, n: int) -> bool:
if n <= 0:
return False
return (math.log10(n) / math.log10(3))%1 == 0
```