Question

1) log_(1/2)(4) + log_3(81)​

Answer 1

Решение.

Применяем свойства логарифмов :

$\bf log_{a}b^{k}=k\cdot log_{a}\, b\ \ ,\ \ log_{a^{k}}\, b=\dfrac{1}{k}\cdot log_{a}\, b\ \ ,\ \ log_{a}\, a=1\ \ ,\\\\a > 0\ ,\ a\ne 1\ ,\ b > 0$

$\bf log_{1/2}\, 4+log_3\, 81=log_{2^{-1}}\, 2^2+log_3\, 3^4=2\cdot (-1)\cdot log_2\, 2+4\cdot log_3\, 3=\\\\\\=-2+4=2$

Answer 2

To simplify the expression, let's use the properties of logarithms.

1. log_(1/2)(4) represents the logarithm of 4 with base 1/2. We can rewrite this as log(4) / log(1/2) using the change of base formula.

The logarithm of 4 with base 1/2 can be written as: log_(1/2)(4) = log(4) / log(1/2)

The logarithm of 4 with base 10 is log(4), and the logarithm of 1/2 with base 10 is log(1/2). Therefore, we can simplify this as: log_(1/2)(4) = log(4) / log(1/2) = log(4) / log(2^(-1)) = log(4) / (-log(2)) = -log(4) / log(2)

2. log_3(81) represents the logarithm of 81 with base 3.

To simplify this, we can express 81 as a power of 3: 81 = 3^4

Therefore, we have: log_3(81) = log_3(3^4) = 4

Now, we can substitute these simplified values back into the original expression:

log_(1/2)(4) + log_3(81) = -log(4) / log(2) + 4

Note that the base of logarithms doesn't affect the sum of the terms, so we can combine the two terms:

• log(4) / log(2) + 4 = 4 - log(4) / log(2)

This is the simplified expression for log_(1/2)(4) + log_3(81).