Question

Решите log^2 3(27x)+log3(x)=-7

Answer 1

Тудет типо три тэ в квадрате+ тэ минус три тэ

Answer 2

To solve the equation log^2₃(27x) + log₃(x) = -7, we can use some logarithmic properties. Let's start by applying those properties step by step:

1. Product Rule of Logarithms: logₐ(b) + logₐ(c) = logₐ(b * c)
2. Power Rule of Logarithms: logₐ(b^m) = m * logₐ(b)

First, we'll use the Product Rule to combine the logarithms:

log³(27x * x) = log³(27x²)

Now, using the Power Rule, we can bring down the exponent:

log³(27x²) = 2 * log³(27x)

Next, we know that logₐ(a) = 1, so:

log³(27) = 1

Now, rewrite 27 as 3^3:

log³(3³) = 1

Now we have:

3 * log³(3) = 1

log³(3) = 1/3

Now, go back to the original equation:

2 * log³(27x) = 2 * log³(3x) = -7

Divide both sides by 2:

log³(3x) = -7/2

Now, using the exponential form of logarithms:

3^(log³(3x)) = 3^(-7/2)

3x = 3^(-7/2)

To simplify the right-hand side:

3^(-7/2) = 1 / 3^(7/2)

Now, rewrite 3^(7/2) as (3^(1/2))^7 = √3^7:

3^(-7/2) = 1 / √(3^7)

Now, we can rationalize the denominator by multiplying the numerator and denominator by √(3^7):

3^(-7/2) = √(3^7) / (3^7)

Now the equation becomes:

3x = √(3^7) / (3^7)

To calculate the right-hand side:

√(3^7) ≈ √(2187) ≈ 46.81

3^7 = 2187

So, the equation becomes:

3x ≈ 46.81 / 2187

Now divide to find the approximate value of x:

x ≈ 0.0214

Therefore, the approximate value of x that satisfies the equation is x ≈ 0.0214.