log 2/x

Logarithms problems.

log 2/x

Postby tonymicro » Wed Apr 29, 2020 2:07 pm

can someone show me how to workout and graph y=log base2 2/x
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Re: log 2/x

Postby Guest » Thu May 21, 2020 1:44 pm

Do you not know the basic rules of logarithms?
log(ab)= log(a)+ log(b) and log(a/b)= log(a)- log(b).

So log(2/x)= log(2)- log(x).

Further, since this logarithm is "base 2", log(2)= 1 so
this is 1- log(x).

The rest involves calculating values.
2^0= 1 so log(1)= 0. With y(x)= log(2/x)= 1- log(x), y(0)= 1- 0= 1.

2^1= 2 so log(2)= 1. y(2)= 1- 1= 0.

2^2= 4 so log(4)= 2. y(4)= 1- 2= -1

2^3= 8 so log(8)= 3. y(8)= 1- 3= -2

For negative x, 2^{-1}= 1/2 so log(1/2)= -1. y(1/2)= 1-(-1)= 2

2^{-2}= 1/4 so log(1/4)= -2. y(1/4)= 1- (-2)= 3.

Mark the points (1/4, 3), (1/2, 2), (0, 1), (2, 0), (4, -1), (8, -2) and draw a smooth curve through them. Do you see that x= 0 will be a vertical asymptote?
Guest
 

Re: log 2/x

Postby Guest » Thu May 21, 2020 1:48 pm

"log(0)= 1" is a typo. It should be log(1)= 0.
Guest
 


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