Help with these two would be grealy appreciated.
1. On the left hand side, you're missing an argument for the log. Nevertheless:
Note that $\displaystyle 2 = \log_5 25$ and so the right hand side is $\displaystyle \log_5 25 + \log_5 2 = \log_5 50$.
The left hand side can also be expressed as a single log. Do it.
Now equate the arguments of the logs on each side and solve for x.
Then the left hand side can be written $\displaystyle \log_5 [(x-5) x]$.
The right hand side can be written $\displaystyle \log_5 50$.
Therefore $\displaystyle (x - 5) x = 50 \Rightarrow x^2 - 5x - 50 = 0 \Rightarrow x = .....$
Warning: Although the quadratic equation has two solutions, only one is valid (why?).
$\displaystyle 2^x \cdot 2^y=16 \iff 2^{x+y}=2^4$
Now we note from here the bases are the same so the exponents must be equal. So we get $\displaystyle x+y=4$ now we multiply it by -2 and add to the one below to get
$\displaystyle 2x+3y=9$
$\displaystyle -2x-2y=-8$
$\displaystyle y=1$
Subbing back in we get
$\displaystyle x=3$