Question

(log4(x))^2-3log4(x)=3^(log3(4)

Answer 1

Условие:

$\bigg( \log_{4}(x) \bigg) {}^{2} - 3 \log_{4}(x) = 3 {}^{ \log_{3}(4) } .$

ОДЗ: х > 0 . Пусть t = log_4(x). По свойству а^(log_a(b))=b. Получим:

$t {}^{2} - 3t = 4. \\ t {}^{2} - 3t - 4 = 0.$

По т. Виета: t=-1; 4.

Обратная замена:

$1) \: \: \: \log_{4}(x) = - 1 \to \: x = 4 {}^{ - 1} = \dfrac{1}{4} = 0.25. \\ 2) \: \: \: \log_{4}(x) = 4 \to \: x = 4 {}^{4} = 256.$

Правила:

$\log_{a}(x) = b \equiv \: x = a {}^{b} .$

Ответ: 0,25 и 256.

Answer 2

To solve the equation, let's break it down step by step.

Given equation: (log4(x))^2 - 3log4(x) = 3^(log3(4))

Let's simplify the right-hand side first: 3^(log3(4))

Since the base of the exponent is 3 and the base of the logarithm is also 3, we can rewrite this expression as: 3^(log3(4)) = 4

Now, let's substitute this value back into the original equation: (log4(x))^2 - 3log4(x) = 4

Let's simplify further by making a substitution: Let y = log4(x)

Now, we have the equation in terms of y: y^2 - 3y = 4

Rearranging the equation: y^2 - 3y - 4 = 0

This is a quadratic equation. To solve it, we can factorize or use the quadratic formula. Factoring the equation, we get: (y - 4)(y + 1) = 0

Setting each factor equal to zero and solving for y: y - 4 = 0 or y + 1 = 0

Solving for y: y = 4 or y = -1

Now, substitute the values of y back in terms of log4(x): log4(x) = 4 or log4(x) = -1

Solving each equation for x: For log4(x) = 4: 4^4 = x x = 256

For log4(x) = -1: 4^(-1) = x 1/4 = x

Therefore, the solutions to the equation are x = 256 and x = 1/4.