Evaluate the given logarithm. log 9 1 81 \log_9\frac{1}{81} lo g 9 81 1

In this problem, we're asked to evaluate the given logarithm. And here, I have log base 9 of one over 81. Now, looking at this right away, I can't see that I can already apply a property. So I'm actually going to need to rewrite this log so that I can apply one of these properties. So So I'm gonna think about what I can do here in order to rewrite this so that one of these properties will work. Now I know that 81 is just 9 squared, so I can rewrite this as a log base 9 of 1 over 9 squared, but it's still not quite in a form that we can use one of these properties. But since this 9 squared is on the bottom of a fraction where the top is just 1 I can actually just rewrite this as log base 9 of 9 to the power of negative 2. Making that exponent negative means that it's just on the bottom of a fraction. So now looking at this I see that my base 9 is the same as this exponent here 9 to the power of negative 2 which will just cancel using our inverse property. So log base 9 and 9 cancel, leaving you with just this negative 2. So my answer here is that log base 9 of 1 over 81 is equal to negative 2. Let me know if you have any questions.

0. Functions

Properties of Logarithms

Calculus

0. Functions

Properties of Logarithms

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Evaluate the given logarithm.
$\log_9\frac{1}{81}$

$81$

$2$

$-2$

$9$

Verified step by step guidance

Recognize that the problem involves evaluating a logarithm: $ \log_9 \left( \frac{1}{81} \right) $.

Recall the property of logarithms that $ \log_b \left( \frac{1}{a} \right) = -\log_b(a) $. This means we can rewrite the expression as $ -\log_9(81) $.

Express 81 as a power of 9. Since $ 81 = 9^2 $, we can substitute this into the logarithm to get $ -\log_9(9^2) $.

Apply the power rule of logarithms, which states $ \log_b(a^c) = c \cdot \log_b(a) $. This simplifies our expression to $ -2 \cdot \log_9(9) $.

Since $ \log_9(9) = 1 $ (because any number to the power of 1 is itself), the expression simplifies to $ -2 \cdot 1 $.