$3^{1 + 2\log_3(y-x)} = 48$ With this problem I have difficulty getting rid of the exponent.
$2\log_5(2y - x - 12) = \log_5(y-x) + \log_5(y + x)$
$3^{1 + 2\log_3(y-x)} = 48$ With this problem I have difficulty getting rid of the exponent.
$2\log_5(2y - x - 12) = \log_5(y-x) + \log_5(y + x)$
Hint: For the first equation, use $3^{a+b}=3^a \cdot 3^b$, then the definition of $\log_3$ is that $3^{\log_3 x}=$ what?
For the second, raise $5$ to the power of each side.
$3^{1+2\log_3(y-x)}=3\cdot 3^{2\log_3(y-x)}=3\cdot (3^{\log_3(y-x)})^2=3(y-x)^2$ using $a^{\log_ax}=x$
$3(y-x)^2=48\implies (y-x)^2=16,y=x\pm 4$
From the 2nd equation, $(2y-x-12)^2=(y-x)(y+x)$ using $\log_x (ab)=\log_x a+\log_x b$ where $x >0,\ne1$
If $y=x+4,x=y-4,\{2y-(y-4)-12\}^2=\{y+y-4\}4$
or $(y-8)^2=4(2y-4),y^2-24y+80=0,y=20$ or $4$ using the solution of quadratic equation.
If $y=20,x=20-4=16;y=4,x=4-4=0$
Similarly, for $y=x-4$