need help with the second one

3 ^{x^2 - 42}   =   1 / 3^x

Note that the term on the right side can be written as 3^-x

So   we have

3 ^{x^2 - 42}  = 3 ^-x

Since the bases are the same, we can solve for the exponents

x^2 - 42 = - x       add x to both sides

x^2 + x - 42 = 0       factor

( x + 7) ( x - 6) =  0

Setting each factor to 0 and solving for x produces  the integer solutions

x = -7    or    x = 6