Question

1) log0,2 125^-3 2) log64 16

Answer 1

Ответ:

1)Не знаю 2)2/3

Пошаговое объяснение:

Answer 2

Ответ:

Пошаговое объяснение:

Answer 3

Let's solve the given logarithmic expressions step by step:

1. log₀.₂(125⁻³):

The notation "log₀.₂" indicates that the logarithm has a base of 0.2. To evaluate this expression, we need to find what power of 0.2 gives us 125⁻³.

125⁻³ can be written as (1/125)³, and 125 is 5³ (5 * 5 * 5).

Therefore, 125⁻³ = (1/5)³.

Now, we need to find the exponent of 0.2 that gives us (1/5)³:

0.2^x = (1/5)³.

To solve for x, take the logarithm of both sides with the same base as 0.2:

log₀.₂(0.2^x) = log₀.₂((1/5)³).

By the logarithm property, the exponent (x) can be brought down in front of the logarithm:

x * log₀.₂(0.2) = 3 * log₀.₂(1/5).

Now, we can find the values of the logarithms using the change of base formula:

log₀.₂(0.2) ≈ log(0.2) / log(0.2).

log₀.₂(1/5) ≈ log(1/5) / log(0.2).

Calculate these values:

log(0.2) ≈ -2.3219, log(1/5) ≈ -0.69897.

Substitute the values back into the equation:

x * (-2.3219) = 3 * (-0.69897).

Now, solve for x:

x ≈ (3 * -0.69897) / -2.3219 ≈ 0.8981.

So, log₀.₂(125⁻³) ≈ 0.8981.

2. log₆₄(16):

This logarithm has a base of 64, which is 2⁶ (2 * 2 * 2 * 2 * 2 * 2).

So, log₆₄(16) = log₂(16).

Since 16 is 2⁴ (2 * 2 * 2 * 2), we have:

log₂(16) = log₂(2⁴).

By the logarithm property, the exponent (x) can be brought down in front of the logarithm:

log₂(2⁴) = 4 * log₂(2).

The logarithm of any number with the base equal to itself is 1, so:

log₂(2) = 1.

Substitute this value back into the equation:

4 * log₂(2) = 4 * 1 = 4.

So, log₆₄(16) = 4.