Help please!

\(\frac{24t^3}{15t^4}*\frac{5t^8}{3t^6}\)

Simplify \(\frac{24t^3}{15t^4}\)

Both 24 and 15 have a common factor of 3

\(\frac{24t^3}{15t^4} = \frac{8t^3}{5t^4}\)

\(\frac{8t^3}{5t^4}*\frac{5t^8}{3t^6}\)

Apply exponent rule: \(\frac{x^a}{x^b} = \) 1/x^b-a

\(\frac{8}{5t}\)

Continue

\(\frac{8}{5t}*\frac{5t^8}{3t^6}\)

Apply exponent rule: \(\frac{x^a}{x^b} = \) x^a-b

\( \frac{5t^2}{3}\)

Continue

\(\frac{8}{5t} * \frac{5t^2}{3}\)

\(\frac{8*5t^2}{5t*3}\)

\(\frac{8t^2}{t*3}\)

\(\frac{8t}{3}\)

Your answer would be \(\frac{8t}{3}\)

Hope this helps ;P

EmeraldWonder, there are a couple of small errors in your work at the end because  \(\frac{8\,*\,5t^2}{5t\,*\,3}\ \neq\ \frac{8t^2}{3}\)

Here is another way to work this problem:

\(\ \phantom{=\quad}\dfrac{24t^3}{15t^4}\cdot \dfrac{5t^8}{3t^6}\)

\(=\quad\dfrac{24\,\cdot\,t^3\,\cdot\,5\,\cdot\,t^8}{15\,\cdot\,t^4\,\cdot\,3\,\cdot\,t^6}\)

\(=\quad\dfrac{120\,\cdot\,t^3\,\cdot\,t^8}{45\,\cdot\,t^4\,\cdot\,t^6}\)          because we can multiply numbers in any order and   24 · 5 = 120   and   15 · 3 = 45

\(=\quad\dfrac{120\,\cdot\,t^{11}}{45\,\cdot\,t^{10}}\)          because   t³ · t⁸  =  t t t · t t t t t t t t  =  t^{(3 + 8)}  =  t¹¹   and   t⁴ · t⁶  =  t^{(4 + 6)}  =  t¹⁰

\(=\quad\dfrac{120\,\cdot\,t^{10}\cdot\,t}{45\,\cdot\,t^{10}}\)          because   t¹¹  =  t t t t t t t t t t t  =  t¹⁰ · t

\(=\quad\dfrac{120\,\cdot\,t}{45}\)          when  t ≠ 0  because we can cancel the common factor of  t¹⁰  in the numerator and denominator.

\(=\quad\dfrac{8t}{3}\)          because   \(\frac{120}{45}\)   reduces to   \(\frac83\)_