the logarithms of 2 to the base 16 =y

solve for y

Question

the logarithms of 2 to the base 16 =y

solve for y

amistre64 · Answer

logB(arg) = exp

log16(2) = y perhaps?  means  16^y = 2

anonymous · Answer

ya i got 16^y=2 but how to you get y

amistre64 · Answer

16^1 = 16
16^0 = 1

so its between 0 and 1 :)

amistre64 · Answer

logs and exponents are inverse; you can try to see if there is a 2^n that equals 16 to help out

amistre64 · Answer

then flip it and see if it works :)

amistre64 · Answer

16^(1/2) = sqrt(16) = 4  so its less than 1/2

amistre64 · Answer

sqrt(4) = 2 tho

sqrt(sqrt(16)) = 16^(1/2)^(1/2) which equals 16^(1/4)

anonymous · Answer

it is..

anonymous · Answer

16^y=2
(4^2)^y=2^1
2y=1
y=0.5

amistre64 · Answer

$$B^{n^m} = B^{n*m}$$

amistre64 · Answer

.5 is close, but not quite

B^(1/2) = sqrt(B)

16^(1/2) ?=? 2

sqrt(16) ?=? 2

          4 ?=? 2  nope

anonymous · Answer

no that is the answer lol.. its in my book

anonymous · Answer

nvm i wrote it wrong

anonymous · Answer

the answer is..

amistre64 · Answer

:)  playing games are we lol

anonymous · Answer

16^y=2
(2^4)^y=2^1
4y=1
y=0.25

amistre64 · Answer

thats correct

anonymous · Answer

yusss!

anonymous · Answer

how about log base x 20=1

amistre64 · Answer

logX(20) = 1  means  X^1 = 20

x = 20 then right?

anonymous · Answer

k good.. checking my work:P thanks!

amistre64 · Answer

logB(B) = 1   is a standard rule of logs

anonymous · Answer

how about log base x x^5=5

amistre64 · Answer

logx(x^5) = 5  ; logB(M^n) = n logB(M)

5 logx(x) = 5  ; logB(B) = 1, so x is all values that are positive

anonymous · Answer

huh?

amistre64 · Answer

plug in any value for x and you get the same results; there is no 1 solution, it is infinite

anonymous · Answer

it will always equal 5?

amistre64 · Answer

barring of course for the fact that log(n<=0) is a nono

amistre64 · Answer

any value of x will work so long as its greater than 0

amistre64 · Answer

log100(100^5) = 5

log2(2^5) = 5

log1/23(1/23^5) = 5

anonymous · Answer

how about..
log base 32 y =2/5

amistre64 · Answer

log32(y) = 2/5

y = 32^(2/5)  then

amistre64 · Answer

logB(y) = x  means  y = B^x

anonymous · Answer

does y = 4?

amistre64 · Answer

dunno, id have to calculator it :)

amistre64 · Answer

yes y=4

anonymous · Answer

i wrote 4 so maybe ill keep it like that..

how about..

log y 81=4/3      :)

amistre64 · Answer

post some new qs so others can get a chance :)  and good luck, youre doing great!!

anonymous · Answer

ahh but you're the best!