Question

If log 3 = .4771, find log (.81)^2 x log(27/10)^2/3 ÷log 9.

Answer 1

Step-by-step explanation:

here u go..................

Answer 2

- 0.05520 is the answer.

Step-by-step explanation:

Given: log 3 = 0.4771

To Find: the value of $log (0.81)^2 \times log(27/10)^{2/3} \div log 9$

Solution:

• Finding the value of $log (0.81)^2 \times log(27/10)^{2/3} \div log 9$ by using logarithmic rules

We have $log (0.81)^2 \times log(27/10)^{2/3} \div log 9$ such that, we can write

$\Rightarrow 2 log (81/100) \times \frac{2}{3} log(27/10) \div log 9 \ \because \ logx^n = nlogx$

$\Rightarrow 2 (log 81 - log 100) \times \frac{2}{3} (log 27 - log10) \div log 9 \ \because \ log(a/b) = log(a) - log(b)$

$\Rightarrow 2 (log 3^4 - log 10^2) \times \frac{2}{3} (log 3^3 - log10) \div log 3^2$

$\Rightarrow 2 (4log 3 - 2log 10) \times \frac{2}{3} (3log 3 - log10) \div 2log 3 \ \because \ logx^n = nlogx$

$\Rightarrow 4 (2log 3 - log 10) \times \frac{2(3log 3 - log10)}{3} \div 2log 3$

$\Rightarrow \frac{4 (2log 3 - log 10) \times (log 3 - log10)}{3log3}$

Now, putting the values log3=0.4771 and log10=1 in the above expression, we get,

$\Rightarrow \frac{4 (2(0.4771) - 1) \times (3 (0.4771) - 1)}{3(0.4771)} = -0.05520$

Hence, the value of $log (0.81)^2 \times log(27/10)^{2/3} \div log 9$ is -0.05520