log concept needed?

Question

if x >y (x,y +ve integers) then can we say log x > log y

kindly provide some logic/examples in support of your answer

thanks

Answer 1

log(3) = 1.099
log(2) = 0.693
log(1) = 0

The graph of y = ln(x) is a monotonic function ie it is always increasing.
dy/dx = 1/x > 0 for x>0
This is proof that ln(x) is an increasing function.
So if ln(x) is increasing then
ln(a) > ln(b) if a>b

**
Look at it another way.
Suppose a > b and
lna = p, lnb = q
a = e^p, b = e^q
Since a>b
a/b > 1
e^p / e^q > 1
e^(p-q) > 1
Thus p-q > 0
p > q
or ln(a) > ln(b)

Answer 2

I guess you mean natural logarithm, right? Well, log is a strictly increasing function on (0, oo). So, whenever 0 < x < y, we have log x < log y, even if x and y are not integers.

The proof that ln x < ln y if 0 < x > y depends on the definition you adopt for the log function.

Answer 3

Dr. D is correct. But I want to give a simpler explanation.

The expression is true because of the following reason:
Let log x = a
log y = b

Writing them in the exponential form :
10^a = x
10^b = y

If x>y, then 10 should be raised to a greater power.

So a > b
or log x > log y

You can apply the same logic for any base.
You may also substantiate your explanation by example.

Hope this helps.

[e-mail]

Answer 4

Yes. The domain of a log is (0,infinity). The function log(x) increases forever. On an interval when a function is increasing, f(a)>f(b) if a>b. The log increases across the entire domain, so log(x)>log(y) given x>y.

Answer 5

Yes.

x > y implies x/y > 1
We know any number > 1 will have log > 0

log(x/y) > 0
log(x) - log(y) > 0
log(x) > log(y)