Question

log

Answer 1

\[\log _{17}17^{y}\]

Answer 2

And how do I get to the desired answer, I'm more confused as to the method.

Answer 3

log17(17) = 1

so the answer seems to be "y"

Answer 4

or to put it another way:

17^? = 17^y

?=y

Answer 5

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?

Answer 6

Y would be 1 in that case?

Answer 7

y would be any real number in this case...

Answer 8

there is no = sign to attribute to any specific value

Answer 9

Do you recall that logs are just exponents?

Answer 10

it just said "Find:"

Answer 11

yeah, put in a different form

Answer 12

B^x = y

logB(B^x) = logB(y)

x = logB(y)

Answer 13

meaning, they're exponents reconfigured to find the y correct?

Answer 14

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?

Answer 15

yes

Answer 16

yes

Answer 17

so, logB(3^2) is just an "exponent" with an exponent..simply put

logB(3^2) = 2 logB(3)

does that make sense?

Answer 18

It makes more sense.

Answer 19

Still shaky, but I'm getting there.

Answer 20

good :) cause making less sense is bad lol

Answer 21

Haahaha

Answer 22

So I need to learn how to reconfigure my exponents then.

Answer 23

in a sense, yes :)

Answer 24

Into logarithmic forms and such. Kinda thinking out loud.

Answer 25

logB(B) = what?

B^? = B ??

B^1 = B right? so,

logB(B) = 1

Answer 26

those b's are bases?

Answer 27

of exponents?

Answer 28

take your log now:

log17(17^y) turns into:

y log17(17) which becomes:

y (1) = y

Answer 29

B = base, yes

Answer 30

Damn dude, you're a mathemagician

Answer 31

lol ..... its the rabbit up my sleeve that gets the credit :)

Answer 32

hi-o

Answer 33

Alright, well i really appreciate the tutelage.

Answer 34

no prob...

Answer 35

i'm sure i'll be back for more haha. peace.