anonymous
  • anonymous
log
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\log _{17}17^{y}\]
anonymous
  • anonymous
And how do I get to the desired answer, I'm more confused as to the method.
amistre64
  • amistre64
log17(17) = 1 so the answer seems to be "y"

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
or to put it another way: 17^? = 17^y ?=y
anonymous
  • anonymous
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?
anonymous
  • anonymous
Y would be 1 in that case?
amistre64
  • amistre64
y would be any real number in this case...
amistre64
  • amistre64
there is no = sign to attribute to any specific value
amistre64
  • amistre64
Do you recall that logs are just exponents?
anonymous
  • anonymous
it just said "Find:"
anonymous
  • anonymous
yeah, put in a different form
amistre64
  • amistre64
B^x = y logB(B^x) = logB(y) x = logB(y)
anonymous
  • anonymous
meaning, they're exponents reconfigured to find the y correct?
amistre64
  • amistre64
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?
amistre64
  • amistre64
yes
anonymous
  • anonymous
yes
amistre64
  • amistre64
so, logB(3^2) is just an "exponent" with an exponent..simply put logB(3^2) = 2 logB(3) does that make sense?
anonymous
  • anonymous
It makes more sense.
anonymous
  • anonymous
Still shaky, but I'm getting there.
amistre64
  • amistre64
good :) cause making less sense is bad lol
anonymous
  • anonymous
Haahaha
anonymous
  • anonymous
So I need to learn how to reconfigure my exponents then.
amistre64
  • amistre64
in a sense, yes :)
anonymous
  • anonymous
Into logarithmic forms and such. Kinda thinking out loud.
amistre64
  • amistre64
logB(B) = what? B^? = B ?? B^1 = B right? so, logB(B) = 1
anonymous
  • anonymous
those b's are bases?
anonymous
  • anonymous
of exponents?
amistre64
  • amistre64
take your log now: log17(17^y) turns into: y log17(17) which becomes: y (1) = y
amistre64
  • amistre64
B = base, yes
anonymous
  • anonymous
Damn dude, you're a mathemagician
amistre64
  • amistre64
lol ..... its the rabbit up my sleeve that gets the credit :)
anonymous
  • anonymous
hi-o
anonymous
  • anonymous
Alright, well i really appreciate the tutelage.
amistre64
  • amistre64
no prob...
anonymous
  • anonymous
i'm sure i'll be back for more haha. peace.

Looking for something else?

Not the answer you are looking for? Search for more explanations.