We're glad you stopped by c tex log 9 frac 1 25 log 9 5 tex. This page is here to walk you through essential details with clear and straightforward explanations. Our goal is to make your learning experience easy, enriching, and enjoyable. Start exploring and find the information you need!
Answer :
To solve the problem [tex]\(\log _9 \frac{1}{25} = \text{Ø} \log _9 5\)[/tex], let's analyze each side of the equation and see if they are equal.
1. Calculate [tex]\(\log_9 \frac{1}{25}\)[/tex]:
Using the properties of logarithms, we can rewrite [tex]\(\log_9 \frac{1}{25}\)[/tex] as:
[tex]\[
\log_9 \frac{1}{25} = \log_9 1 - \log_9 25
\][/tex]
Based on the logarithmic identity, [tex]\(\log_b 1 = 0\)[/tex] for any base [tex]\(b\)[/tex], and knowing that:
[tex]\[
\log_9 25 \approx 2 \times \log_9 5
\][/tex]
We simplify the expression:
[tex]\[
\log_9 \frac{1}{25} = 0 - 2 \times \log_9 5 = -2 \times \log_9 5
\][/tex]
Numerically, this value is approximately [tex]\(-1.46497\)[/tex].
2. Evaluate [tex]\(\log_9 5\)[/tex]:
Now, let's consider the term [tex]\(\log_9 5\)[/tex]. The value of [tex]\(\log_9 5\)[/tex] can be computed, but for the purpose here, it is given to be approximately [tex]\(0.73249\)[/tex].
3. Comparison:
We have:
[tex]\[
\log_9 \frac{1}{25} \approx -1.46497
\][/tex]
[tex]\[
2 \times \log_9 5 \approx 1.46497
\][/tex]
We see that [tex]\(\log_9 \frac{1}{25}\)[/tex] does not equal [tex]\(\log_9 5\)[/tex], reaffirming that [tex]\(-2 \times \log_9 5\)[/tex] and [tex]\(\log_9 5\)[/tex] are not equal.
Thus, the initial statement [tex]\(\log _9 \frac{1}{25} = \text{Ø} \log _9 5\)[/tex] is false. The steps reveal that the correct relationship involves a factor of [tex]\(-2\)[/tex] instead, not simply Ø (which implies equality).
1. Calculate [tex]\(\log_9 \frac{1}{25}\)[/tex]:
Using the properties of logarithms, we can rewrite [tex]\(\log_9 \frac{1}{25}\)[/tex] as:
[tex]\[
\log_9 \frac{1}{25} = \log_9 1 - \log_9 25
\][/tex]
Based on the logarithmic identity, [tex]\(\log_b 1 = 0\)[/tex] for any base [tex]\(b\)[/tex], and knowing that:
[tex]\[
\log_9 25 \approx 2 \times \log_9 5
\][/tex]
We simplify the expression:
[tex]\[
\log_9 \frac{1}{25} = 0 - 2 \times \log_9 5 = -2 \times \log_9 5
\][/tex]
Numerically, this value is approximately [tex]\(-1.46497\)[/tex].
2. Evaluate [tex]\(\log_9 5\)[/tex]:
Now, let's consider the term [tex]\(\log_9 5\)[/tex]. The value of [tex]\(\log_9 5\)[/tex] can be computed, but for the purpose here, it is given to be approximately [tex]\(0.73249\)[/tex].
3. Comparison:
We have:
[tex]\[
\log_9 \frac{1}{25} \approx -1.46497
\][/tex]
[tex]\[
2 \times \log_9 5 \approx 1.46497
\][/tex]
We see that [tex]\(\log_9 \frac{1}{25}\)[/tex] does not equal [tex]\(\log_9 5\)[/tex], reaffirming that [tex]\(-2 \times \log_9 5\)[/tex] and [tex]\(\log_9 5\)[/tex] are not equal.
Thus, the initial statement [tex]\(\log _9 \frac{1}{25} = \text{Ø} \log _9 5\)[/tex] is false. The steps reveal that the correct relationship involves a factor of [tex]\(-2\)[/tex] instead, not simply Ø (which implies equality).
We appreciate you taking the time to read c tex log 9 frac 1 25 log 9 5 tex. We hope the insights shared have been helpful in deepening your understanding of the topic. Don't hesitate to browse our website for more valuable and informative content!
- Why do authors use plot complications in stories A To resolve all a story s conflicts at once B To increase suspense and interest C
- For one month Siera calculated her hometown s average high temperature in degrees Fahrenheit She wants to convert that temperature from degrees Fahrenheit to degrees.
Rewritten by : Batagu
