Question

Log2√2 1/8=? Log1/√2в 5 степени 1/4=? Помогите пожалуйста!

Answer 1

Log2 1/8 = log2 2^-3 = -3 log2 2 = -3 *1 = -3

Answer 2

Calculation 1: log2√2 + 1/8

To calculate log2√2 + 1/8, we need to break it down step by step.

First, let's simplify the expression log2√2. The square root of 2 (√2) can be written as 2^(1/2). Therefore, log2√2 can be rewritten as log2(2^(1/2)).

Using the logarithmic property logb(b^x) = x, we can simplify log2(2^(1/2)) to (1/2).

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

log2√2 + 1/8 = (1/2) + 1/8

To add these fractions, we need a common denominator. The least common denominator (LCD) of 2 and 8 is 8. Therefore, we can rewrite the expression as:

(1/2) + 1/8 = 4/8 + 1/8 = 5/8

So, log2√2 + 1/8 is equal to 5/8.

Calculation 2: log(1/√2)^(5/4)

To calculate log(1/√2)^(5/4), we need to simplify the expression.

First, let's simplify the expression inside the logarithm. The square root of 2 (√2) can be written as 2^(1/2). Therefore, 1/√2 can be written as 1/(2^(1/2)).

Using the exponentiation property (a^m)^n = a^(m*n), we can simplify 1/(2^(1/2)) to 2^(-1/2).

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

log(1/√2)^(5/4) = log(2^(-1/2))^(5/4)

Using the exponentiation property logb(b^x) = x, we can simplify log(2^(-1/2))^(5/4) to (-1/2) * (5/4).

Multiplying these fractions, we get:

(-1/2) * (5/4) = -5/8

So, log(1/√2)^(5/4) is equal to -5/8.

I hope this helps! Let me know if you have any further questions.