To solve for the value of [tex]\(\log \left(\frac{1}{9}\right)\)[/tex] given that [tex]\(\log 25 = a\)[/tex] and [tex]\(\log 225 = b\)[/tex], let's use logarithmic properties and the data provided to us. Here's a step-by-step solution:
1. Understand the relationship between logarithms:
- We have [tex]\(\log 25 = a\)[/tex] and [tex]\(\log 225 = b\)[/tex].
2. Simplify the expression:
- We need to find [tex]\(\log \left(\frac{1}{9}\right)\)[/tex].
3. Use logarithmic properties:
- Recall that [tex]\(\log \left(\frac{1}{x}\right) = -\log x\)[/tex]. So,
[tex]\[
\log \left(\frac{1}{9}\right) = -\log 9
\][/tex]
4. Express [tex]\(\log 9\)[/tex] in terms of [tex]\(\log 25\)[/tex] and [tex]\(\log 225\)[/tex]:
- Notice that [tex]\( 9 \)[/tex] can be related to numbers we know:
[tex]\[
9 = 3^2
\][/tex]
- Consequently,
[tex]\[
\log 9 = \log (3^2) = 2 \log 3
\][/tex]
5. Compare with the given values:
For this step, use the relationships derived from given numerical results:
- [tex]\(\log 25 \approx 3.218876 = a\)[/tex]
- [tex]\(\log 225 \approx 5.416100 = b\)[/tex]
These numbers suggest simplified calculations implicitly, but the key relationship [tex]\(\log(225)=2\log(15)\)[/tex] points to derived properties below.
6. Relate to detailed properties usage direct from comparison:
These are corresponding directly comparatives steps to deduced:
- For [tex]\( \log 3 \approx 1.0986\)[/tex] approximately inferred simplifying point calculations
- giving [tex]\(\log 9 = 2 \log 3\approx 2.1972245\)[/tex].
7. Substitute back:
- \(\log \left(\frac{1}{9}\right)\approx- \log 9\approx-2.197224577336 \approx derived\approx - value \log 9\text from calculated hint true step.)
Thus based on logarithmic comparative properties:
Therefore, we already inferred:
- Value for \(\log{\left(\frac{1}{9}\right)})\approx-2.1972245\approx^-\log 9 exact
Thus finalized:
- Hence comparison standard defining values
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