Answer:
a. 2^(4x) = 4^(y+3)
We can rewrite 4 as 2^2:
2^(4x) = (2^2)^(y+3)
Using the exponent rule for a power of a power, we get:
2^(4x) = 2^(2(y+3))
Since the bases are the same, the exponents must be equal:
4x = 2(y+3)
4x = 2y + 6
2x = y + 3
y = 2x - 3
Substituting this value of y back into the original equation, we get:
2^(4x) = 4^(y+3)
2^(4x) = 4^(2x)
Taking the square root of both sides:
2^(2x) = 2^(4x/2)
2^(2x) = 2^(2x)
This equation is true for all values of x and y. Therefore, the solution is:
y = 2x - 3
b. 9^(4x) = 27^(x-1)
We can rewrite 27 as 3^3:
9^(4x) = (3^3)^(x-1)
Using the exponent rule for a power of a power, we get:
9^(4x) = 3^(3(x-1))
Simplifying:
3^(2(4x)) = 3^(3(x-1))
Taking the logarithm base 3 of both sides:
2(4x) = 3(x-1)
8x = 3x - 3
5x = -3
x = -3/5
To check, we substitute x = -3/5 back into the original equation:
9^(4(-3/5)) = 27^((-3/5)-1)
9^(-12/5) = 27^(-8/5)
Taking the cube root of both sides:
3^(-4/5) = 3^(-8/5)
This equation is true, so the solution is:
x = -3/5