$$ \log_{3}24 - 3\log_{3}5\times \log_{5}2$$
What I can get is:
$$ \log_3{24} - \log_3{5^3} \times \log_{5}2$$
Change of base rule to get it all in base 3:
$$ \log_5{2} = \frac{\log_3{2}}{\log_3{5}} $$
Now I have:
$$\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}$$
How to continue from here?
Just write $$\frac{\ln(24)}{\ln(3)}-\frac{3\ln(5)}{\ln(3)}\cdot \frac{\ln(2)}{\ln(5)}$$ and $$\ln(24)=\ln(3\cdot 2^3)=\ln(3)+3\ln(2)$$ and with this we get $$\frac{\ln(3)+3\ln(2)-3\ln(2)}{\ln(3)}=1$$
$$\begin{align}\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}&=\log_3{24} - \frac{3\log_3{5}\times \log_3{2}}{\log_3{5}}\\ &=\log_3{24} - 3\log_3 2\\ \end{align}$$ Now you can use the exponent law again on the second term, and then the additive law. Can you finish from here?
Continuing from what you have, \begin{align} \log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}} &= \log_3{24} - \frac{3\cdot\log_3{5}\cdot \log_3{2}}{\log_3{5}} \\ &=(\log_3{3} + \log_3{8}) - 3\cdot \log_3{2} \\ &= 1 + 3 \log_3 2 - 3 \log_3 2 \\ &= 1. \end{align}
In questions like these, it's often a good idea to reduce exponents as much as possible: you started by converting $3 \log_3 5$ to $\log_3 (5^3)$, but the former expression is simpler to work with. You can see this further in how expanding $\log_3 24$ was a useful step to finish.
