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   $\frac83$_