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 = 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 = ±√ 2^x = ±√ 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 = ± 2^x = ± 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 = 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 = ±√
2^x = ±√
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 = ±
2^x = ±
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).