Question 93116: How do i evaluate the following:
log3 81
log4 256
log2 (1/32)
log5 1
log3 (-1)
Please help
thanks
Lucy Found 2 solutions by stanbon, venugopalramana:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! How do i evaluate the following?
Comment: log is just another name for "exponent" or "power"
--------------
log3 81
Let the answer be "x"
Then x= log3 81
So x is the exponent of 3 that gives you 81
3^x = 81
3^x = 3^4
x=4
log3 81 = 4
---------------------
log4 256
Similarly
x = log4 256
4^x = 256
4^x = 4^4
x = 4
--------------
log2 (1/32)
2^x = (1/32)
2^x = 2^-5
x = -5
---------------
log5 1
5^x = 1
5^x = 5^0
x=0
-------------
log3 (-1)
3^x = -1
No such x exists.
No solution.
===================
Cheers,
Stan H.
--------------
You can put this solution on YOUR website! How do i evaluate the following:
log3 81
WRITE THE NUMBER AS A POWER OF BASE.THAT POWER IS THE LOG OF THE NUMBER TO THE BASE.
HERE NUMBER = N =81
BASE = B =3
WE KNOW THAT
3*3*3*3=3^4=81........P=4
HENCE
LOG(81) TO BASE 3 = 4
[LOGARITHM] = P=4
log4 256
N=256.......B=4............4^4=256.....P=5
LOGARITHM =4
log3 (-1)
N=-1....B=3.........WE HAVE NO POWER P WHICH MAKES 3^P=-1
SO IT DOES NOT EXIST...
INFACT LOG OF 0 OR NEGATIVE NUMBER DO NOT EXIST
WE CAN HAVE LOGS FOR POSITIVE NUMBERS ONLY.
Please help
thanks
Lucy
(