Sure, let's evaluate each of these logarithms step by step.
1. Evaluate [tex]\(\log_{12}(144)\)[/tex]:
To find [tex]\(\log_{12}(144)\)[/tex], we use the definition of a logarithm, which states that if [tex]\(b^\alpha = x\)[/tex], then [tex]\(\log_b(x) = \alpha\)[/tex].
Here, [tex]\(12^\alpha = 144\)[/tex]. We need to find [tex]\(\alpha\)[/tex].
We notice that [tex]\(144\)[/tex] can be written as [tex]\(12^2\)[/tex] because [tex]\(144 = 12 \times 12 = 12^2\)[/tex].
Hence, [tex]\(\alpha = 2\)[/tex].
Therefore, [tex]\(\log_{12}(144) = 2\)[/tex].
2. Evaluate [tex]\(\log_{15}(1)\)[/tex]:
For any base [tex]\(b\)[/tex], if the argument of the logarithm is [tex]\(1\)[/tex], then the value is always [tex]\(0\)[/tex]. This is because any number raised to the power of [tex]\(0\)[/tex] is [tex]\(1\)[/tex].
Therefore, [tex]\(\log_{15}(1) = 0\)[/tex].
3. Evaluate [tex]\(\log_{3}\left(\frac{1}{81}\right)\)[/tex]:
We need to find [tex]\(\log_{3}\left(\frac{1}{81}\right)\)[/tex].
Notice that [tex]\(81\)[/tex] is a power of [tex]\(3\)[/tex]: [tex]\(81 = 3^4\)[/tex].
Therefore, [tex]\(\frac{1}{81} = 81^{-1} = \left(3^4\right)^{-1} = 3^{-4}\)[/tex].
So, [tex]\(\log_{3}(3^{-4}) = -4\)[/tex] because [tex]\(\log_b(b^k) = k\)[/tex].
Therefore, [tex]\(\log_{3}\left(\frac{1}{81}\right) = -4\)[/tex].
4. Evaluate [tex]\(\log(0.00001)\)[/tex]:
When the base is not specified in a logarithm, it is usually assumed to be base 10. So, [tex]\(\log(0.00001)\)[/tex] essentially refers to [tex]\(\log_{10}(0.00001)\)[/tex].
We can write [tex]\(0.00001\)[/tex] as a power of 10:
[tex]\[
0.00001 = 10^{-5}
\][/tex]
So, [tex]\(\log_{10}(10^{-5}) = -5\)[/tex] because [tex]\(\log_b(b^k) = k\)[/tex].
Therefore, [tex]\(\log(0.00001) = -5\)[/tex].
Summarizing all the results:
[tex]\[
\log_{12}(144) = 2
\][/tex]
[tex]\[
\log_{15}(1) = 0
\][/tex]
[tex]\[
\log_{3}\left(\frac{1}{81}\right) = -4
\][/tex]
[tex]\[
\log(0.00001) = -5
\][/tex]
