Issue With Algebra of Logarithms

  • #1

Main Question or Discussion Point

I was just doing a homework problem that involved logarithms.

I noticed that order of operations matters when applying logarithm rules.

I'll use a different example from my homework problem to illustrate what I'm talking about.

ln(5*2^3) does not equal 3*(ln(5*2))

Apparently you have to do ln5+3*ln2 - you have to separate out the products before multiplying the second term by the exponent! It kind of makes sense but I can't say exactly why this is how it has to be.

Can anyone confirm that what I have said is true and explain why?
 

Answers and Replies

  • #2
symbolipoint
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Logarithms as numbers, are exponents, and they follow the rules of exponents.

Not know which base you want to work with, but log(5*2^3)
log(5)+log(2^3)
log(5)+3*log(2)
and your book would explain how and why better than I can right now.
 
  • #3
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Well consider the simpler example of ##5*2^3## vs ##(5*2)^3##
 
  • #4
symbolipoint
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Well consider the simpler example of ##5*2^3## vs ##(5*2)^3##
That is 40.
If want log(40), this is if base ten,
log(4*10)
log(4)+log(10)
log(4)+1
0.60206+1
1.60206, approximately

This also means that 10^1.60206=40, approximately
101.60206=40
 
  • #5
Well consider the simpler example of ##5*2^3## vs ##(5*2)^3##
Yes, that does put things in perspective a bit!
 
  • #6
symbolipoint
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This can also help reenforce things a little more.

23*25=2*2*2*2*2*2*2*2
2(3+5)
28


The EXPONENTS are LOGARITHMS. Here, the base was 2.
 
  • #7
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ln(5*2^3) does not equal 3*(ln(5*2))
There's an order of operations at play here that has nothing to do with logarithms.

In the expression 5 * 2^3, the exponent operator takes precedence over the multiplication, making this expression equal to 5 * 8. The mnemonic is PEMDAS (or BEDMAS), with each emphasizing that exponentiation takes precedence over multiplication or division. (he latter two operations have the same precedence.)

Going back to the log problem, this means that
##\log(5 \cdot 2^3) = \log(5 \cdot 8) = \log(5) + \log(8)##

If the log expression had been ##\log((5\cdot 2)^3)##, then you could have used the log property you're using to get ##3\log(5 \cdot 2) = 3\log(10)##
 
  • #8
There's an order of operations at play here that has nothing to do with logarithms.

In the expression 5 * 2^3, the exponent operator takes precedence over the multiplication, making this expression equal to 5 * 8. The mnemonic is PEMDAS (or BEDMAS), with each emphasizing that exponentiation takes precedence over multiplication or division. (he latter two operations have the same precedence.)

Going back to the log problem, this means that
##\log(5 \cdot 2^3) = \log(5 \cdot 8) = \log(5) + \log(8)##

If the log expression had been ##\log((5\cdot 2)^3)##, then you could have used the log property you're using to get ##3\log(5 \cdot 2) = 3\log(10)##
Okay, that helps a lot. I like to be able to do the algebra following the rules for it without having to memorize special exceptions. Here, this is the way to look at it from the perspective of PEMDAS. Thanks!
 
  • #9
symbolipoint
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Learning the rules for logarithms should not be very long in time. Really only a few of them are there. Applying them in steps will become part of your nature.
 

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