Question

1/2log корень из 21 9+log21 49 log6 5*log5 8+log6 27

Answer 1

$\frac{1}{2}log_{ \sqrt{21} } 9+log_{21}49= log_{ \sqrt{21} } \sqrt{9} +log_{21}49 =log_{21}9+log_{21}49=$ $log_{21}9*49 = log_{21}441 = log_{21}21^2=2*log_{21}21=2*1=2$


$log_65*log_58+log_627 = log_65* \frac{log_68}{log_65}+log_627=log_68+log_627=$ $log_68*27=log_6216=log_66^3=3*1=3$

Answer 2

To simplify the expression, let's start by using logarithm properties. We have:

1. $\frac{1}{2}\log(\sqrt{21})$
2. $9 + \log_{21}49$
3. $\log_6 5 \cdot \log_5 8$
4. $\log_6 27$

Now, let's simplify each part step by step:

1. $\frac{1}{2}\log(\sqrt{21})$ The square root of 21 is $\sqrt{21} = \sqrt{7 \cdot 3} = \sqrt{7} \cdot \sqrt{3}$ Taking the logarithm: $\log(\sqrt{21}) = \log(\sqrt{7} \cdot \sqrt{3}) = \log(\sqrt{7}) + \log(\sqrt{3}) = \frac{1}{2} \log(7) + \frac{1}{2} \log(3)$ Therefore, $\frac{1}{2}\log(\sqrt{21}) = \frac{1}{2} \cdot \frac{1}{2} \log(7) + \frac{1}{2} \cdot \frac{1}{2} \log(3) = \frac{1}{4} \log(7) + \frac{1}{4} \log(3)$

2. $9 + \log_{21}49$ Recall that $\log_a b$ is the exponent to which we must raise $a$ to get $b$. So, $\log_{21}49 = \frac{\log 49}{\log 21}$ $49$ can be written as $7^2$, so $\log_{21}49 = \frac{\log(7^2)}{\log 21} = \frac{2\log 7}{\log 21}$ Now, $9 + \frac{2\log 7}{\log 21} = \frac{9\log 21 + 2\log 7}{\log 21}$

3. $\log_6 5 \cdot \log_5 8$ We can use the property that $\log_a b \cdot \log_b c = \log_a c$ So, $\log_6 5 \cdot \log_5 8 = \log_6 8$