To solve the equation [16^x] - 6*[4^x] + 5 = 0, we can rewrite it using the properties of exponents.
Let's start by expressing 16 and 4 as powers of 2: 16 = 2^4 4 = 2^2
Substitute these values into the equation: [2^(4x)] - 6*[2^(2x)] + 5 = 0
Next, use the properties of exponents when raising a power to another power: 2^(4x) - 6*2^(2x) + 5 = 0
Let's simplify the equation:
Let y = 2^x, y^2 = 2^(2x), y^4 = 2^(4x)
Now, our equation becomes: y^4 - 6y^2 + 5 = 0
This can be factored as: (y^2 - 5)(y^2 - 1) = 0
So, y^2 = 5 or y^2 = 1
This results in two cases:
1) y^2 = 5: y = ±√5 2^x = ±√5 x = log2(√5) or x = log2(-√5) - Note: This case is not valid, as the logarithm of a negative number is not defined for real numbers.
2) y^2 = 1: y = ±1 2^x = ±1 x = 0 or x = log2(-1) - Note: This case is not valid as well, as the logarithm of a negative number is not defined for real numbers.
Therefore, the only valid solution is x = log2(√5).
To solve the equation [16^x] - 6*[4^x] + 5 = 0, we can rewrite it using the properties of exponents.
Let's start by expressing 16 and 4 as powers of 2:
16 = 2^4
4 = 2^2
Substitute these values into the equation:
[2^(4x)] - 6*[2^(2x)] + 5 = 0
Next, use the properties of exponents when raising a power to another power:
2^(4x) - 6*2^(2x) + 5 = 0
Let's simplify the equation:
Let y = 2^x, y^2 = 2^(2x), y^4 = 2^(4x)
Now, our equation becomes:
y^4 - 6y^2 + 5 = 0
This can be factored as:
(y^2 - 5)(y^2 - 1) = 0
So, y^2 = 5 or y^2 = 1
This results in two cases:
1) y^2 = 5:
y = ±√5
2^x = ±√5
x = log2(√5) or x = log2(-√5) - Note: This case is not valid, as the logarithm of a negative number is not defined for real numbers.
2) y^2 = 1:
y = ±1
2^x = ±1
x = 0 or x = log2(-1) - Note: This case is not valid as well, as the logarithm of a negative number is not defined for real numbers.
Therefore, the only valid solution is x = log2(√5).