Question

http://prnt.sc/cllv2i please help

Answer 1

Part A: since anything to the 0 power is 1, x = 0.

Part B: Since 6^0 is already 1, and 1 to any power is 1, x can be any set of whole numbers.

Answer 2

\[6^{2x}=1=6^0,2x=0,x=\frac{ 0 }{ 2 }=0\]
\[\left( 6^0 \right)^x=1,1^x=1,minimum~ value ~of~x=0 \]
and other values are all natural numbers.

Answer 3

So for both x =0
@sshayer

Answer 4

no for second all whole numbers or 0 and positive integers
or 0 and all natural numbers
all three answers are same.