3^√128 = 4^2x 3^x^2-4x-05 = 81√3

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вопрос

Let's solve each equation separately:

1) Solve for x in the equation 3^(√128) = 4^(2x):
First, simplify both sides of the equation:
3^(√128) = 3^(√(642)) = 3^(√(4^3 2)) = 3^(√4^3) 3^(√2) = 4^3 3^(√2) = 4 4^2 3^(√2) = 4 16 3^(√2) = 64 * 3^(√2).

4^(2x) = 4^(2log4(3^(√2))) = 4^(log4(3^(√2))^2) = (3^(√2))^2 = 9 3^2 = 81.

Therefore, we have 64 * 3^(√2) = 81.
Now, we can solve for √2:
3^(√2) = 81/64 = 1.265625.
√2 ≈ log3(1.265625) = 0.267075.

2) Solve for x in the equation 3^x^2 -4x - 0.5 = 81√3:
First, simplify the right side of the equation:
81√3 = 81 3^(1/2) = 81 √3 = 81 * 1.732 = 140.292.

Now, rewrite the equation as:
3^x^2 -4x - 0.5 = 140.292,
3^x^2 - 4x = 140.792.

There isn't an analytical way to solve this equation, and we would need to use numerical methods (such as Newton's method) to find the value(s) of x that satisfy the equation.