Certainly! Let's evaluate [tex]\( \log_3 \left(\frac{1}{27}\right) \)[/tex] step-by-step.
1. Rewrite [tex]\( \frac{1}{27} \)[/tex] in exponential form:
Let's first express [tex]\( \frac{1}{27} \)[/tex] as a power of 3. We know that [tex]\( 27 = 3^3 \)[/tex], so:
[tex]\[
\frac{1}{27} = 3^{-3}
\][/tex]
2. Use the logarithm property:
Recall the logarithm property [tex]\( \log_b (a^c) = c \cdot \log_b (a) \)[/tex]. We can apply this property here:
[tex]\[
\log_3 \left(3^{-3}\right) = -3 \cdot \log_3 (3)
\][/tex]
3. Simplify using the definition of logarithms:
We know that [tex]\( \log_3 (3) \)[/tex] is the logarithm of 3 to the base 3, which equals 1:
[tex]\[
\log_3 (3) = 1
\][/tex]
4. Calculate the final value:
Substitute this back into our equation:
[tex]\[
-3 \cdot \log_3 (3) = -3 \cdot 1 = -3
\][/tex]
Therefore, the value of [tex]\( \log_3 \left(\frac{1}{27}\right) \)[/tex] is:
[tex]\[
\log_3 \left(\frac{1}{27}\right) = -3
\][/tex]
For completeness, let's provide the intermediate steps in logarithmic form for clarity:
- The natural logarithm of the value and the base would be:
[tex]\[
\ln \left(\frac{1}{27}\right) \approx -3.295836866004329
\][/tex]
[tex]\[
\ln (3) \approx 1.0986122886681098
\][/tex]
- Applying the change of base formula:
[tex]\[
\log_3 \left(\frac{1}{27}\right) = \frac{\ln \left(\frac{1}{27}\right)}{\ln (3)} \approx \frac{-3.295836866004329}{1.0986122886681098} \approx -3
\][/tex]
Hence, confirming that:
[tex]\[
\log_3 \left(\frac{1}{27}\right) = -3
\][/tex]
