Using numerical analysis and the Power Rule of Logarithms, logx^k=klogx

Question

, how can log(1/100)
be rewritten?(1 point)
Responses

2log−10
2 logarithm negative 10

log−20
logarithm negative 20

10log−2
10 logarithm negative 2

−2log10

Answer 1

To rewrite $\log \left(\frac{1}{100}\right)$ using the properties of logarithms, we can use the fact that $\log \left(\frac{a}{b}\right)=\log (a)-\log (b)$.

1. We can rewrite $\log \left(\frac{1}{100}\right)$ as: $$\log (1)-\log (100)$$

2. We know that $\log (1)=0$, so we have: $$0-\log (100)=-\log (100)$$

3. Now, we can express $\log (100)$ as $\log ({10}^2)$ and apply the Power Rule of Logarithms: $$-\log (100)=-\log ({10}^2)=-2\log (10)$$

4. Therefore, we can rewrite $\log \left(\frac{1}{100}\right)$ as: $$-2\log (10)$$

So the final answer is:

−2log10