Question

Log5 log32 2 =? срочно​

Answer 1

Ответ:

-1

Пошаговое объяснение:

$log_{5}( log_{32}(2) ) = log_{5}( {5}^{ - 1} ) = - 1$

Answer 2

Ответ:

-1.

Пошаговое объяснение:

log5 log32 2 = log5 (log2^5 2) = log5 (1/5log2 2) = log5 (1/5•1) = log5 5^(-1) = -1•log5 5 = -1•1 = -1.

Answer 3

To solve the equation log₅(log₃₂(2)), we need to use the properties of logarithms.

First, let's start with the innermost logarithm, log₃₂(2). This logarithm represents the exponent to which we need to raise 32 to obtain 2. In other words, we need to find the power that 32 needs to be raised to in order to get 2.

32¹ = 32, 32² = 1024, and 32³ = 32768. From this, we can see that 32² = 1024 is the closest power of 32 that is less than 2.

Therefore, log₃₂(2) is approximately equal to 0.5.

Now, we can substitute this value back into the original equation:

log₅(log₃₂(2)) = log₅(0.5).

Since log₅(0.5) represents the exponent to which we need to raise 5 to obtain 0.5, we can rewrite it as:

5^x = 0.5.

To solve for x, we need to rewrite the equation using exponential notation:

x = log₅(0.5).

By evaluating this expression, we find:

x ≈ -0.4307.

Therefore, log₅(log₃₂(2)) is approximately equal to -0.4307.