How do you condense #Ln 3 − 2 ln(8 + 4) #?

Answer 1

#-ln(48)#

We have to remember two different log rules: Rule 1: #alogb = log(b^a) # Rule 2: #log(a) + log(b) = log(ab)# (Note: Rule 1 is actually a specific case of Rule 2 if #a=b# and you do it a bunch of times)

From that, we can simplify the original: #ln(3) - 2ln(8+4) # #ln(3) - 2ln(12) # (by addition) #ln(3) + ln(12^(-2)) # (by rule 1) #ln(3 * 12^(-2)) # (by rule 2) #ln(3/144) # (by definition) #ln(1/48) = - ln(48) # our final answer.

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Answer 2

Elias Ackerman

To condense the expression Ln 3 − 2 ln(8 + 4), you can use the properties of logarithms:

Ln 3 − 2 ln(8 + 4) = Ln 3 − Ln(8 + 4)^2
Ln 3 − Ln(8 + 4)^2 = Ln 3 − Ln(12)^2
Ln 3 − Ln(12)^2 = Ln 3 − Ln(144)
Ln 3 − Ln(144) = Ln(3/144)
Ln(3/144) = Ln(1/48)

So, Ln 3 − 2 ln(8 + 4) is condensed to Ln(1/48).

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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.