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anonymous
 5 years ago
log
anonymous
 5 years ago
log

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And how do I get to the desired answer, I'm more confused as to the method.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0log17(17) = 1 so the answer seems to be "y"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or to put it another way: 17^? = 17^y ?=y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's what the book says. Did you get that by putting it in exponential form and then seeing to what multiple the 17 was?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Y would be 1 in that case?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y would be any real number in this case...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there is no = sign to attribute to any specific value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Do you recall that logs are just exponents?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, put in a different form

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0B^x = y logB(B^x) = logB(y) x = logB(y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0meaning, they're exponents reconfigured to find the y correct?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when we have an "exponent" with an exponent; what can we do with those exponents? multiply them together right? b^2^4 = b^8 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so, logB(3^2) is just an "exponent" with an exponent..simply put logB(3^2) = 2 logB(3) does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Still shaky, but I'm getting there.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good :) cause making less sense is bad lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I need to learn how to reconfigure my exponents then.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Into logarithmic forms and such. Kinda thinking out loud.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0logB(B) = what? B^? = B ?? B^1 = B right? so, logB(B) = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0take your log now: log17(17^y) turns into: y log17(17) which becomes: y (1) = y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Damn dude, you're a mathemagician

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol ..... its the rabbit up my sleeve that gets the credit :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, well i really appreciate the tutelage.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm sure i'll be back for more haha. peace.
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