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Evaluate the following logarithms.

[tex]\[ \log_{12} 144 = \square \][/tex]

If [tex]\( b^a = x \)[/tex], then [tex]\( \log_b x = a \)[/tex], where [tex]\( b \ \textgreater \ 0 \)[/tex] and [tex]\( b \neq 1 \)[/tex].

[tex]\[ \log_{15} 1 = \square \][/tex]

[tex]\[ \log_3\left(\frac{1}{81}\right) = \square \][/tex]

[tex]\[ \log 0.00001 = \square \][/tex]

Sagot :

GinnyAnsweridn
Alright, let's evaluate the given logarithms step-by-step.

### 1. Calculating [tex]\(\log_{12} 144\)[/tex]

We are asked to find [tex]\(\log_{12} 144\)[/tex]. The logarithm [tex]\(\log_b x\)[/tex] asks the question: "To what power must we raise [tex]\(b\)[/tex] to get [tex]\(x\)[/tex]?
Here, we need to find the power [tex]\(a\)[/tex] such that:
[tex]\[ 12^a = 144 \][/tex]

By calculation, we find that:
[tex]\[ 12^2 = 144 \][/tex]

Thus:
[tex]\[ \log_{12} 144 = 2 \][/tex]
So, [tex]\(\log_{12} 144 = 2.0\)[/tex].

### 2. Calculating [tex]\(\log_{15} 1\)[/tex]

We are asked to find [tex]\(\log_{15} 1\)[/tex]. The logarithm [tex]\(\log_b x\)[/tex] in this case is asking for:
[tex]\[ b^a = x \][/tex]
Here, we need to find the power [tex]\(a\)[/tex] such that:
[tex]\[ 15^a = 1 \][/tex]

By definition, any number to the power of 0 is 1:
[tex]\[ 15^0 = 1 \][/tex]

Thus:
[tex]\[ \log_{15} 1 = 0 \][/tex]
So, [tex]\(\log_{15} 1 = 0.0\)[/tex].

### 3. Calculating [tex]\(\log_3 \left(\frac{1}{81}\right)\)[/tex]

We are asked to find [tex]\(\log_3 \left(\frac{1}{81}\right)\)[/tex]. This logarithm is asking for:
[tex]\[ 3^a = \frac{1}{81} \][/tex]

We know that:
[tex]\[ 81 = 3^4 \][/tex]
Thus:
[tex]\[ \frac{1}{81} = 3^{-4} \][/tex]

Therefore:
[tex]\[ 3^a = 3^{-4} \][/tex]
[tex]\[ a = -4 \][/tex]

So:
[tex]\[ \log_3 \left(\frac{1}{81}\right) = -4 \][/tex]
Thus, [tex]\(\log_3 \left(\frac{1}{81}\right) = -4.0\)[/tex].

### 4. Calculating [tex]\(\log 0.00001\)[/tex]

In this case, we are finding the common logarithm (base 10) of [tex]\(0.00001\)[/tex]. The logarithm [tex]\(\log_b x\)[/tex] in this instance is asking for:
[tex]\[ 10^a = 0.00001 \][/tex]

We know that:
[tex]\[ 0.00001 = 10^{-5} \][/tex]

Therefore:
[tex]\[ 10^a = 10^{-5} \][/tex]
[tex]\[ a = -5 \][/tex]

So:
[tex]\[ \log 0.00001 = -5 \][/tex]
Thus, [tex]\(\log 0.00001 = -5.0\)[/tex].

To summarize:

[tex]\[ \log_{12} 144 = 2.0 \][/tex]

[tex]\[ \log_{15} 1 = 0.0 \][/tex]

[tex]\[ \log_3 \left(\frac{1}{81}\right) = -4.0 \][/tex]

[tex]\[ \log 0.00001 = -5.0 \][/tex]

These are the evaluated logarithms.
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