Please help me solve: 1) 3^x+4 = 27 2)4^3k = 64 3) -81 = -3^2x+8?

Question

4) 3^2 - 2x = 1


Thank you

Answer 1

I'm guessing at the question since you didn't give any parentheses.

1)
3^(x+4) = 27
3^(x + 4) = 3^3
Now that both are written as 3 to some exponent, just set the exponents equal.
x + 4 = 3
x = -1

2)
4^(3k) = 64
4^(3k) = 4^3
3k = 3
k = 1

3)
-81 = -3^(2x+8)
-(3^4) = -[3^(2x + 8)]
4 = 2x + 8
-4 = 2x
-2 = x

4)
3^(2 - 2x) = 1
3^(2 - 2x) = 3^0 (anything to the 0 power = 1)
2 - 2x = 0
2 = 2x
1 = x

Answer 2

1) do you mean 3^(x+4) or 3^x + 4? if the 1st, then
3^(x+4) = 27 = 3³
x + 4 = 3
x = -1

2) 4^(3k) = 64 = 4³
3k = 3
k = 1

3) -81 = -[ 3^(2x+8)]
3^4 = 3^(2x+8)
4 = 2x + 8
-4 = 2x
-2 = x

4) 3^(2 - 2x) = 1 = 3^0
2 - 2x = 0
2 = 2x
x = 1

be more careful with () in future.