Question

express the following fraction in simplest form, only using positive exponents.\\(
\frac{-5v^{-10}}{-2(v^{-2})^3}
\\)

Explanation:

Step1: Simplify the negative signs and coefficients

The fraction has \(-5\) in the numerator and \(-2\) in the denominator. The product of two negative numbers is positive, so \(\frac{-5}{-2}=\frac{5}{2}\).

Step2: Simplify the exponent in the denominator

Using the power of a power rule \((a^m)^n = a^{m\times n}\), for \((v^{-2})^3\), we have \(v^{-2\times3}=v^{-6}\). So the denominator becomes \(-2v^{-6}\), but we already handled the negative signs in Step 1, so now we focus on the exponents of \(v\).

Step3: Simplify the exponents of \(v\)

Using the rule for dividing exponents \(a^m\div a^n=a^{m - n}\), we have \(v^{-10}\div v^{-6}=v^{-10-(-6)}=v^{-10 + 6}=v^{-4}\). But we need positive exponents, and using the rule \(a^{-n}=\frac{1}{a^n}\), \(v^{-4}=\frac{1}{v^4}\) is incorrect here. Wait, no, let's re - do the exponent calculation. Wait, the original fraction is \(\frac{-5v^{-10}}{-2(v^{-2})^3}\). After simplifying the coefficients \(\frac{-5}{-2}=\frac{5}{2}\), and simplifying the denominator's exponent: \((v^{-2})^3 = v^{-6}\), so now we have \(\frac{5v^{-10}}{2v^{-6}}\). When dividing with exponents, \(a^m\div a^n=a^{m - n}\), so \(v^{-10}\div v^{-6}=v^{-10-(-6)}=v^{-4}\). But we want positive exponents, so \(v^{-4}=\frac{1}{v^4}\) is wrong? Wait, no, wait: \(\frac{v^m}{v^n}=v^{m - n}\), so if we have \(\frac{v^{-10}}{v^{-6}}\), that's \(v^{-10-(-6)}=v^{-4}\), but to make the exponent positive, we can rewrite \(v^{-4}\) as \(\frac{1}{v^4}\) only if we move it to the denominator, but wait, no, let's check again. Wait, the fraction is \(\frac{5v^{-10}}{2v^{-6}}\), we can rewrite this as \(\frac{5}{2}\times v^{-10}\times v^{6}\) (because dividing by \(v^{-6}\) is multiplying by \(v^{6}\)). Then \(v^{-10}\times v^{6}=v^{-10 + 6}=v^{-4}\), and then \(\frac{5}{2}v^{-4}=\frac{5}{2}\times\frac{1}{v^{4}}\)? No, that's not right. Wait, no, I made a mistake. Wait, the rule is \(a^{-n}=\frac{1}{a^{n}}\), so \(v^{-4}=\frac{1}{v^{4}}\), but that would be \(\frac{5}{2v^{4}}\), but that's not correct. Wait, no, wait, let's start over.

Original expression: \(\frac{-5v^{-10}}{-2(v^{-2})^3}\)

First, simplify the denominator's exponent: \((v^{-2})^3=v^{-2\times3}=v^{-6}\)

So the expression becomes \(\frac{-5v^{-10}}{-2v^{-6}}\)

Now, simplify the coefficients: \(\frac{-5}{-2}=\frac{5}{2}\)

Now, simplify the \(v\) terms: \(\frac{v^{-10}}{v^{-6}}=v^{-10-(-6)}=v^{-4}\)

But we want positive exponents. Recall that \(a^{-n}=\frac{1}{a^{n}}\), so \(v^{-4}=\frac{1}{v^{4}}\), but that would give \(\frac{5}{2}\times\frac{1}{v^{4}}\)? No, that's wrong. Wait, no, the correct rule is \(\frac{a^m}{a^n}=a^{m - n}\), so if \(m=-10\) and \(n = - 6\), then \(m - n=-10-(-6)=-4\), so \(v^{-4}\). But to write with positive exponents, we can use the rule \(a^{-n}=\frac{1}{a^{n}}\), so \(v^{-4}=\frac{1}{v^{4}}\), but that would be \(\frac{5}{2v^{4}}\), but that's incorrect. Wait, no, I think I messed up the sign when moving the term. Wait, actually, \(\frac{v^{-10}}{v^{-6}}=v^{-10 + 6}=v^{-4}\), and \(\frac{5}{2}v^{-4}=\frac{5}{2}\times\frac{1}{v^{4}}\) is wrong. Wait, no, no, the correct way is: \(\frac{v^m}{v^n}=v^{m - n}\), so if we have \(\frac{v^{-10}}{v^{-6}}\), it's \(v^{-10-(-6)}=v^{-4}\), and to make the exponent positive, we can rewrite \(v^{-4}\) as \(\frac{1}{v^{4}}\) only if we are moving it from the numerator to the denominator or vice - versa. Wait, no, the numerator has \(v^{-10}\) and the denominator has \(v^{-6}\), so \(\frac{v^{-10}}{v^{-6}}=\frac{1}{v^{10}}\times v^{6}=\frac{v^{6}}{v^{10}}=\frac{1}{v^{4}}\). Ah, there we go. So \(\frac{v^{-10}}{v^{-6}}=\frac{1}{v^{10}}\times v^{6}=\frac{v^{6}}{v^{10}}=\frac{1}{v^{4}}\). Then multiply by the coefficient \(\frac{5}{2}\), so we get \(\frac{5}{2v^{4}}\)? No, that's not right. Wait, no, I think I made a mistake in the sign of the exponents. Wait, let's use the rule \(\frac{a^m}{a^n}=a^{m - n}\) correctly. \(v^{-10}\div v^{-6}=v^{-10-(-6)}=v^{-4}\), and \(v^{-4}=\frac{1}{v^{4}}\), so \(\frac{5}{2}\times\frac{1}{v^{4}}=\frac{5}{2v^{4}}\)? But that seems wrong. Wait, no, wait, let's check with numbers. Let \(v = 2\).

Original expression: \(\frac{-5\times(2)^{-10}}{-2\times((2)^{-2})^{3}}\)

First, calculate denominator: \((2^{-2})^3=2^{-6}\), so denominator is \(-2\times2^{-6}=-2^{-5}\) (since \(a^m\times a^n=a^{m + n}\), \(-2\times2^{-6}=-2^{1+( - 6)}=-2^{-5}\))

Numerator is \(-5\times2^{-10}\)

So the fraction is \(\frac{-5\times2^{-10}}{-2^{-5}}=\frac{5\times2^{-10}}{2^{-5}}=5\times2^{-10-(-5)}=5\times2^{-5}=\frac{5}{32}\)

Now, let's calculate \(\frac{5}{2v^{4}}\) when \(v = 2\): \(\frac{5}{2\times(2)^4}=\frac{5}{2\times16}=\frac{5}{32}\), which matches. So that's correct. Wait, but earlier I thought it was wrong, but it's correct.

Wait, but let's do it again step by step:

1. Simplify the negative signs: \(\frac{-5}{-2}=\frac{5}{2}\)
2. Simplify the exponent in the denominator: \((v^{-2})^3 = v^{-6}\) (using \((a^m)^n=a^{mn}\))
3. Now, the expression is \(\frac{5v^{-10}}{2v^{-6}}\)
4. Use the rule \(\frac{a^m}{a^n}=a^{m - n}\) for the \(v\) terms: \(v^{-10-(-6)}=v^{-4}\)
5. Rewrite \(v^{-4}\) with a positive exponent: \(v^{-4}=\frac{1}{v^{4}}\)
6. Multiply by the coefficient: \(\frac{5}{2}\times\frac{1}{v^{4}}=\frac{5}{2v^{4}}\)

Answer:

\(\frac{5}{2v^{4}}\)