anonymous
  • anonymous
Fill in the missing values to make the equations true. log_4(7)+log_4(5)=log_4(?) log_8(11)-log_8(?)=log_8(11/5) 2log_9(3)=log_9(?) is the first one 12? is the second one 5? the third 3/2?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I think adding logs means multiplying their arguments inside the log. In that case it wouldn't be 12 for the first equation but...?
anonymous
  • anonymous
The second one is right: subtracting logs is like dividing the arguments.
anonymous
  • anonymous
the first one you multiply:) thanks

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anonymous
  • anonymous
And in the third case: multiplication outside the log becomes potentiacion inside it. So rather than saying 3 divided by 2 you'd say three ... 2. Let me know if that doesn't help...
anonymous
  • anonymous
Oups, sorry for my horrible orthography. :-D
campbell_st
  • campbell_st
(1) log(a) + log(b) = log(a x b) the missing part is 7 x5 or 35 (2) log(a - log(b) = log(a/b) the missing part is 5 (3) log(a^b) = blog(a) the missing part 3^2 or 9 which would give an answer of 1 (4)
anonymous
  • anonymous
I'm remembering this like so: If you make a list of operations with increasing "strength" or "power" then addition/subtraction would be the "weakest", multiplication would be "middle" and exponentiation would be "strongest". When you go from outside a log to inside a log then you have to increase the "strength" one step: addition/subtraction outside log becomes multiplication/division inside. And multiplication/division outside becomes exponentiation inside. If this seems just complicated then forget it. Just my personal way to remember it.

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