Answer:
Step-by-step explanation:
Hello, please consider the following.
Question 1.
[tex]\dfrac{3^{2x+1}}{3^{3x-4}\cdot 3^{6-7x}}=27^x\\\\<=> 3^{2x+1}\cdot 3^{-3x+4}\cdot 3^{-6+7x}=3^{2x+1-3x+4-6+7x}=(3^3)^x=3^{3x}\\\\<=> 2x+1-3x+4-6+7x=3x\\\\<=> 6x-1=3x\\\\<=> 3x=1\\\\<=> \boxed{x=\dfrac{1}{3}}[/tex]
Question2.
[tex]2^{x+y}=8=2^3 <=>x+y=3\\\\3^{x-y}=1=3^0<=>x-y=0[/tex]
So, it gives (by adding the two equations) 2x = 3
[tex]\boxed{x=\dfrac{3}{2} \ \ and \ \ y = x = \dfrac{3}{2} }[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
