Question

rewrite using a single positive exponent.
$6^{4} \cdot 6^{-9}$

Explanation:

Step1: Apply exponent product rule

When multiplying exponents with the same base, we add the exponents: \(a^m \cdot a^n = a^{m + n}\). So for \(6^{4} \cdot 6^{-9}\), we have \(6^{4 + (-9)}\).

Step2: Simplify the exponent

Calculate \(4 + (-9)\): \(4 - 9 = -5\), so we get \(6^{-5}\).

Step3: Convert to positive exponent

Using the rule \(a^{-n}=\frac{1}{a^{n}}\), we can rewrite \(6^{-5}\) as \(\frac{1}{6^{5}}\), but we want a single positive exponent in the numerator. Wait, actually, we can also use the rule in reverse: \(a^m \cdot a^n = a^{m + n}\), and if we have a negative exponent, we can think of it as moving the base to the denominator or numerator. But another way: \(a^m \cdot a^n = a^{m + n}\), so \(6^{4} \cdot 6^{-9}=6^{4 - 9}=6^{-5}=\frac{1}{6^{5}}\), but we can also express it as \(6^{-5}=\frac{1}{6^{5}}\), but if we want to write it with a single positive exponent in the form where the base is in the denominator with positive exponent, or we can use the rule \(a^m\div a^n=a^{m - n}\), so \(6^{4}\cdot6^{-9}=6^{4}\div6^{9}=\frac{6^{4}}{6^{9}}\), but we can also write it as \(6^{-5}=\frac{1}{6^{5}}\). Wait, maybe I made a mistake. Wait, the problem says "rewrite using a single positive exponent". Wait, actually, when we have \(a^m \cdot a^n = a^{m + n}\), so \(6^{4} \cdot 6^{-9}=6^{4 - 9}=6^{-5}\), and to make the exponent positive, we can use the reciprocal: \(6^{-5}=\frac{1}{6^{5}}\), but that's a fraction. Wait, maybe the problem allows the base to be in the denominator with a positive exponent, or maybe I misread. Wait, no, the problem says "rewrite using a single positive exponent". Wait, actually, the exponent rules: \(a^m \cdot a^n = a^{m + n}\), so \(6^{4} \cdot 6^{-9}=6^{4 + (-9)}=6^{-5}\), and \(a^{-n}=\frac{1}{a^{n}}\), so \(6^{-5}=\frac{1}{6^{5}}\), which has a single positive exponent \(5\) in the denominator. Alternatively, if we consider that maybe the problem wants it in the form \(\frac{1}{6^{5}}\) or maybe there's a mistake in my approach. Wait, no, let's check again. The problem is \(6^{4} \cdot 6^{-9}\). Using \(a^m \cdot a^n = a^{m + n}\), so \(4 + (-9) = -5\), so \(6^{-5}\). Then, to write with a single positive exponent, we use \(a^{-n}=\frac{1}{a^{n}}\), so \(6^{-5}=\frac{1}{6^{5}}\). So the answer is \(\frac{1}{6^{5}}\) or \(6^{-5}\) is negative, but we need positive exponent, so \(\frac{1}{6^{5}}\) which has exponent \(5\) (positive) in the denominator.

Wait, maybe I messed up. Let's do it again.

Step1: Use the product of powers rule

For any non - zero number \(a\) and integers \(m\) and \(n\), \(a^{m}\cdot a^{n}=a^{m + n}\). Here, \(a = 6\), \(m = 4\) and \(n=-9\). So \(6^{4}\cdot6^{-9}=6^{4+( - 9)}\)

Step2: Calculate the exponent

\(4+( - 9)=4 - 9=-5\), so we have \(6^{-5}\)

Step3: Convert negative exponent to positive

By the definition of negative exponents, \(a^{-n}=\frac{1}{a^{n}}\) for \(a
eq0\) and integer \(n\). So \(6^{-5}=\frac{1}{6^{5}}\)

Answer:

\(\frac{1}{6^{5}}\) (or we can also write it as \(6^{-5}=\frac{1}{6^{5}}\), and since the problem asks for a single positive exponent, \(\frac{1}{6^{5}}\) has a positive exponent \(5\) in the denominator)