Question

express the following fraction in simplest form, only using positive exponents.\\(\frac{15d^{-9}t^{4}}{(5d^{-3}t)^{-2}}\\)

Explanation:

Step1: Apply the negative exponent rule to the denominator

The negative exponent rule states that \((a^m)^{-n}=a^{-mn}\) and \((ab)^{-n}=a^{-n}b^{-n}\). So for \((5d^{-3}t)^{-2}\), we have:
\[
(5d^{-3}t)^{-2}=5^{-2}d^{(-3)\times(-2)}t^{-2}=5^{-2}d^{6}t^{-2}
\]
Now the expression becomes \(\frac{15d^{-9}t^{4}}{5^{-2}d^{6}t^{-2}}\)

Step2: Use the rule for dividing exponents with the same base (\(\frac{a^m}{a^n}=a^{m - n}\)) and for coefficients

First, handle the coefficients: \(\frac{15}{5^{-2}}\). Since \(5^{-2}=\frac{1}{5^{2}}=\frac{1}{25}\), then \(\frac{15}{\frac{1}{25}} = 15\times25=375\)
For the \(d\) terms: \(\frac{d^{-9}}{d^{6}}=d^{-9 - 6}=d^{-15}=\frac{1}{d^{15}}\) (but we want positive exponents, so we'll move it to the denominator later, or we can use the rule \(\frac{a^m}{a^n}=a^{m - n}\) and then handle negative exponents by reciprocating)
For the \(t\) terms: \(\frac{t^{4}}{t^{-2}}=t^{4-(-2)}=t^{6}\)

Wait, actually, when we have \(\frac{15d^{-9}t^{4}}{5^{-2}d^{6}t^{-2}}\), we can rewrite it as \(15\times5^{2}\times d^{-9 - 6}\times t^{4-(-2)}\) (because \(\frac{a}{b^{-n}}=a\times b^{n}\) and \(\frac{a^{m}}{a^{n}}=a^{m - n}\))

So \(15\times25 = 375\), \(d^{-15}\), \(t^{6}\)

Now, since \(d^{-15}=\frac{1}{d^{15}}\), but we can also use the rule of exponents when dividing: \(\frac{a^m}{a^n}=a^{m - n}\), and for negative exponents, \(a^{-n}=\frac{1}{a^{n}}\)

Alternatively, let's do it step by step with exponent rules:

First, simplify the denominator \((5d^{-3}t)^{-2}\) using \((ab)^n=a^n b^n\) and \((a^m)^n=a^{mn}\):

\((5d^{-3}t)^{-2}=5^{-2}(d^{-3})^{-2}t^{-2}=5^{-2}d^{6}t^{-2}\) (because \((a^m)^n=a^{mn}\), so \((d^{-3})^{-2}=d^{6}\))

Now the expression is \(\frac{15d^{-9}t^{4}}{5^{-2}d^{6}t^{-2}}\)

Now, use the rule for dividing terms with the same base: \(\frac{a^m}{a^n}=a^{m - n}\) and \(\frac{a}{b^{-n}}=a\times b^{n}\)

For the coefficient: \(\frac{15}{5^{-2}}=15\times5^{2}=15\times25 = 375\)

For \(d\): \(\frac{d^{-9}}{d^{6}}=d^{-9 - 6}=d^{-15}=\frac{1}{d^{15}}\) (but we can also write it as \(d^{-15}\) and then, since we want positive exponents, we can move it to the denominator, but wait, no—wait, actually, when we have \(d^{-15}\), it's \(\frac{1}{d^{15}}\), but let's check the signs again.

Wait, maybe a better approach:

Recall that \(\frac{1}{(x)^{-n}}=x^{n}\), so \(\frac{1}{(5d^{-3}t)^{-2}}=(5d^{-3}t)^{2}\)

Ah! That's a better rule: \(\frac{1}{a^{-n}}=a^{n}\). So the original expression \(\frac{15d^{-9}t^{4}}{(5d^{-3}t)^{-2}}\) is equal to \(15d^{-9}t^{4}\times(5d^{-3}t)^{2}\) (because dividing by a negative exponent is multiplying by the positive exponent)

Now, expand \((5d^{-3}t)^{2}=5^{2}(d^{-3})^{2}t^{2}=25d^{-6}t^{2}\)

Now the expression is \(15d^{-9}t^{4}\times25d^{-6}t^{2}\)

Multiply the coefficients: \(15\times25 = 375\)

Multiply the \(d\) terms: \(d^{-9}\times d^{-6}=d^{-9 + (-6)}=d^{-15}=\frac{1}{d^{15}}\) (no, wait, when multiplying exponents with the same base, we add the exponents: \(a^m\times a^n=a^{m + n}\))

So \(d^{-9}\times d^{-6}=d^{-15}\), and \(t^{4}\times t^{2}=t^{6}\)

Now, \(d^{-15}=\frac{1}{d^{15}}\), so the expression is \(375\times\frac{1}{d^{15}}\times t^{6}=\frac{375t^{6}}{d^{15}}\)

Wait, but let's check with another method. Let's use the exponent rules for division:

\(\frac{15d^{-9}t^{4}}{(5d^{-3}t)^{-2}}=15d^{-9}t^{4}\times(5d^{-3}t)^{2}\) (because \(\frac{1}{a^{-n}}=a^{n}\))

Then \((5d^{-3}t)^{2}=25d^{-6}t^{2}\)

So \(15\times25 = 375\), \(d^{-9}\times d^{-6}=d^{-15}\), \(t^{4}\times t^{2}=t^{6}\)

So \(375d^{-15}t^{6}=\frac{375t^{6}}{d^{15}}\) (since \(d^{-15…

Answer:

\(\frac{375t^{6}}{d^{15}}\)