Question

1) log3 18 + log3 (1\18) 2) 2log2 1\4 - 3log1\3 27

Answer 1

1)

$log_{3}18+log_{3}\frac{1}{18}=log_{3}(18\cdot \frac{1}{18}) =log_{3}1=0$

2)

$2log_{2}\frac{1}{4} -3log_{\frac{1}{3} }27=2log_{2}2^{-2} -3log_{\frac{1}{3} }(\frac{1}{3})^{-3} =2\cdot (-2)log_{2}2-3\cdot (-3)log_{\frac{1}{3} }\frac{1}{3}=\\\\=-4+9=5$

Answer 2

Let's simplify each of the given expressions step by step:

1. $\log_3 18 + \log_3 \left(\frac{1}{18}\right)$:

Using the properties of logarithms, we can combine these logarithms using the addition rule: $\log_b (a) + \log_b (c) = \log_b (a \cdot c)$.

So, applying this rule to the given expression: $\log_3 18 + \log_3 \left(\frac{1}{18}\right) = \log_3 (18 \cdot \frac{1}{18}) = \log_3 1 = 0$

2. $2 \log_2 \left(\frac{1}{4}\right) - 3 \log_{\frac{1}{3}} 27$:

Again, using the properties of logarithms, we can simplify this expression using the following rules:

• $\log_b (a^n) = n \log_b (a)$
• $\log_b \left(\frac{a}{c}\right) = \log_b (a) - \log_b (c)$

Applying these rules to the given expression: $2 \log_2 \left(\frac{1}{4}\right) - 3 \log_{\frac{1}{3}} 27 = 2 \cdot (-2) \log_2 2 - 3 \cdot (-3) \log_{\frac{1}{3}} 3^3$

Simplifying further: $-4 \log_2 2 + 9 \log_{\frac{1}{3}} 3 = -4 \cdot 1 + 9 \cdot 1 = -4 + 9 = 5$

So, the simplified values of the given expressions are:

1. $\log_3 18 + \log_3 \left(\frac{1}{18}\right) = 0$
2. $2 \log_2 \left(\frac{1}{4}\right) - 3 \log_{\frac{1}{3}} 27 = 5$