Question

Log2(1000)+log2(2)-log2(20)+log5(5)-lg(10)

Answer 1

Ответ:

$log_21000+log_22-log_220+log_55-lg10=log_2\dfrac{1000\cdot 2}{20}+1-1=\\\\\\=log_2100=log_210^2=2\, log_210=2\, log_2(2\cdot 5)=2\, (1+log_25)$

Answer 2

To simplify the given expression:

log2(1000) + log2(2) - log2(20) + log5(5) - lg(10)

Let's break it down step by step:

1. log2(1000) can be simplified to log2(2^3), because 2^3 = 8. So, log2(1000) = 3 * log2(2).

2. log2(2) is simply 1, as log of a number to the base of itself is always 1.

3. log2(20) can be simplified to log2(2^2 * 5), because 20 = 2^2 * 5. Using logarithm properties, this becomes 2 * log2(2) + log2(5).

4. log5(5) is equal to 1, as log of a number to the base of itself is always 1.

5. lg(10) refers to the base 10 logarithm of 10, which is also 1.

Now, substitute these values back into the original expression:

3 * log2(2) + 1 - (2 * log2(2) + log2(5)) + 1 - 1

Now, simplify further:

3 - 2 - 1 + 1 - 1

Finally, perform the arithmetic:

0

So, the simplified expression is 0.