Just a bit confused about how to evaluate the following
$$\log_3 8\times \log_5 9\times \log_2 5$$
What I have done so far:
I have used the change of base rule to change each log to base $3$, so I ended up with this after cancelling: $$\log_3 8\cdot \log_3 9 \cdot \frac{1}{\log_32}$$
First of all, I'm not even sure that I have made the right decision in changing to base 3, does it matter what you change to? Second of all, I'm just unsure where to go from here.
To systematically attack such questions, use $\log_x y = \frac{\log y}{ \log x}$ i.e. $$\log_3 8 = \frac{\ln 8}{\ln 3}, \quad \log_5 9 = \frac{\ln 9}{\ln 5}, \quad \log_2 5 = \frac{\ln 5}{\ln 2}$$ multiply and simplify $$\log_3 8 \times \log_5 9 \times \log_2 5 = \frac{\ln 8}{\ln 3} \times \frac{\ln 9}{\ln 5} \times \frac{\ln 5}{\ln 2}\\ =\frac{\ln 8 \times \ln 9}{\ln 3 \times \ln 2} =\frac{\ln 8}{\ln 2} \times \frac{\ln 9}{\ln 3} = 3\times2 = 6$$
Notice:
$$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$ $$\ln(a^b)=b\ln(a)$$
So:
$$\log_3(8)\cdot\log_5(9)\cdot\log_2(5)=\frac{\ln(8)}{\ln(3)}\cdot\frac{\ln(9)}{\ln(5)}\cdot\frac{\ln(5)}{\ln(2)}=\frac{\ln(8)}{\ln(3)}\cdot\frac{\ln(9)}{1}\cdot\frac{1}{\ln(2)}=\frac{\ln(8)}{\ln(3)}\cdot\frac{\ln(9)}{\ln(2)}=$$ $$\frac{\ln(8)}{\ln(3)}\cdot\frac{2\ln(3)}{\ln(2)}=\frac{\ln(8)}{1}\cdot\frac{2}{\ln(2)}=\frac{3\ln(2)}{1}\cdot\frac{2}{\ln(2)}=\frac{3\ln(2)}{\ln(2)}=3\cdot2=6$$
