# If a rough approximation for ln(5) is 1.609 how do you use this approximation and differentials to approximate ln(128/25)?

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To approximate ln(128/25) using the given approximation for ln(5) as 1.609:

- Recognize that ln(128/25) = ln(128) - ln(25).
- Calculate ln(128) using ln(5) as a base: ln(128) ≈ ln(5) + ln(25) ≈ 1.609 + ln(25).
- Determine ln(25) using ln(5) as a base: ln(25) = 2 * ln(5) ≈ 2 * 1.609.
- Substitute the values into the equation: ln(128/25) ≈ 1.609 + 2 * 1.609.
- Calculate the result: ln(128/25) ≈ 1.609 + 3.218 ≈ 4.827.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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