Question

Question: What will be the value of log3 (1/9) + log 9 81 ?

Answer 1

Answer:

SOLUTION: find the value of the expression log3 81 - log3(1/9) 4 -(-2) = 4 + 2 or 6. So, the answer is 6.

Answer 2

0 is the value of $log_3 (1/9) + log_9(81)$.

Given:

A logarithmic expression $log_3 (1/9) + log_9(81)$.

To Find:

The value of the given logarithmic expression.

Solution:

Write $\frac{1}{9}=\frac{1}{3^2}$ and $81=9^2$ in the given expression.

$log_3 (\frac{1}{3^2}) + log_9(9^2)$

Use the logarithmic property $log_aa^2=2log_aa$ , where a is an integer, in the above-obtained expression.

$log_3 (\frac{1}{3^2}) + 2log_9(9)$

Use the logarithmic property $log_aa=1$  where a is an integer, in the above-obtained expression.

$log_3 (\frac{1}{3^2}) + 2\cdot 1=log_3 (\frac{1}{3^2})+2$

Write $log_3 (\frac{1}{3^2})$ as $log_3 (3^{-2})$ in the obtained expression.

$log_3 (\frac{1}{3^2}) + 2\cdot 1=log_3(3^{-2}})+2$

Again use the logarithmic property $log_aa^2=2log_aa$ where a is an integer, in the obtained expression.

$-2log_33+2$

Now use the logarithmic property $log_aa=1$ where a is an integer, in the obtained expression.

$-2\cdot 1+2=-2+2\\=0$

Thus, 0 is the value of $log_3 (1/9) + log_9(81)$.