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

Answer 1
To approximate #ln(128/25)# using linear approximation and/or differentials we need a number near #128/25# whose #ln# we know.
Clearly #128/25# is somewhat near #125/25#
and #125/25=5# whose #ln# we were given.
The difference between #ln(128/25)# and #ln(5)# is approximately equal to the differential of #y=lnx#
#dy=1/x dx#
To approximate near #5#, we use #dy = 1/5 dx= 1/5 (x-5)#
With #x=128/25#, #x-5=3/25=12/100=0.12#
and #dy=1/2(0.12)=0.024#
#ln(128/25)=ln(125/25)+ Delta y#
#ln(128/25) ~~ ln(125/25)+ dy~~1.609+0.024=1.633#
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Answer 2

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

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

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.

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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