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And how do I get to the desired answer, I'm more confused as to the method.
log17(17) = 1 so the answer seems to be "y"
or to put it another way: 17^? = 17^y ?=y
Yeah, 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?
Y would be 1 in that case?
y would be any real number in this case...
there is no = sign to attribute to any specific value
Do you recall that logs are just exponents?
it just said "Find:"
yeah, put in a different form
B^x = y logB(B^x) = logB(y) x = logB(y)
meaning, they're exponents reconfigured to find the y correct?
when we have an "exponent" with an exponent; what can we do with those exponents? multiply them together right? b^2^4 = b^8 right?
so, logB(3^2) is just an "exponent" with an exponent..simply put logB(3^2) = 2 logB(3) does that make sense?
It makes more sense.
Still shaky, but I'm getting there.
good :) cause making less sense is bad lol
So I need to learn how to reconfigure my exponents then.
in a sense, yes :)
Into logarithmic forms and such. Kinda thinking out loud.
logB(B) = what? B^? = B ?? B^1 = B right? so, logB(B) = 1
those b's are bases?
take your log now: log17(17^y) turns into: y log17(17) which becomes: y (1) = y
B = base, yes
Damn dude, you're a mathemagician
lol ..... its the rabbit up my sleeve that gets the credit :)
Alright, well i really appreciate the tutelage.
i'm sure i'll be back for more haha. peace.