Our test asked to simplify $(\log_9 2 + \log_9 4)\log_2 (3)$.
I simplified the first parenthesis to be $\log_9 (8)$.
So, now I have $\log_9 (8) \cdot \log_2 (3)$ and I can change to base $10$ and get, $$\frac{\log 8}{\log 9} \cdot \frac{\log 3}{\log 2}$$ However, the problem shows that this should without a calculator simplify to $1.5$. Without using a calculator, I don't see how I'm supposed to get $3/2$ from the multiplication of these $2$ logs. Any thoughts?
$$(\log_{9}2+\log_9{4})\log_{2}3$$ this equal to: $$\log_{9}8\cdot \log_{2}3=\frac{3\log2}{2\log3}\cdot\frac{\log3}{\log2}=\frac{3}{2}$$
Hint: $\log_{10} 8 = \log_{10} 2^3 = 3\log_{10} 2$ and $\log_{10} 9 = \log_{10} 3^2 = 2\log_{10} 3$.
Hint. It's a bit confusing with your notation, but you can use the fact that
$$ \log_9 8 \times \log_2 3 = \log_9 8^{\log_2 3} $$
and observe that
$$ 8 = 2^3 $$
It will also be useful to know that (in this context)
$$ (a^b)^c = (a^c)^b $$
$$(\log_{9}2+\log_9{4})\log_{2}3$$ $$=(\log_{9}2+\log_9{2^2})\log_{2}3$$ $$=(\log_{9}2+2*\log_9{2})\log_{2}3$$ $$=(3*\log_9{2})*\log_{2}3$$ $$=3*\log_9{2}*\log_{2}3$$ $$=3*\log_9{3}$$ $$=3*\log_9{9^{1/2}}$$ $$={3/2}*\log_9{9}$$ $$={3/2}$$
