If a = ln22, B = ln33, C = ln55, the order from small to large is______ ．

If a = ln22, B = ln33, C = ln55, the order from small to large is______ ．

∵ a = ln22, B = ln33, C = ln55, ∵ B-A = ln33-ln22 = 2ln3 − 3ln26 = ln9 − ln86 ＞ 0a-c = ln22 − ln55 = 5ln2 − 2ln510 = ln32 − ln2510 ＞ 0, that is, B ＞ a, a ＞ C  C ＜ a ＜ B, so the answer is: C ＜ a ＜ B

Compare the size of LN2 / 2, Ln3 / 3, LN5 / 5

The denominator and numerator of LN2 / 2 are multiplied by 3 to be 3ln2 / 6
3ln2 = ln8,8 is the third power of 2,
Ln8 / 6
And so on, Ln3 / 3 denominator is multiplied by 2,
Result: ln9 / 6
The denominator of two numbers is the same,
So Ln3 / 3 is bigger
The following analogy can determine the size of three numbers

It is known that a = ln 3 / 3, B = LN 2 / 2, C = ln 5 / 5

A = (LN2) / 2 = ln (radical 2),
B = (Ln3) / 3 = ln (root 3 of degree 3),
C = (LN5) / 5 = ln (quintic root 5)
Root 5 < root 2 < root 3
So: C

ln1-ln2=-ln2
How to simplify? Is it like this, ln1-ln2 = ln (1 / 2) = LN2 ^ - 1 = - LN2,

ln1-ln2=ln(1/2)=ln2^-1=-ln2
Your solution is absolutely correct
The formula is
ln a^(-b)=(-b)ln a

Why is ln2-ln1 equal to LN2?
Why is ln2-li1 / 2D equal to 2ln2?

Because ln1 = 0!

LN2 - ln1 = LN2, the calculation process and the calculation method of this kind of problem, it is better to have a formula

lna-lnb=ln(a/b)…

How much is ln0, ln1 and LN2

The function value of function f (x) = [x] represents the calculation of the largest integer not exceeding x [ln1] + [LN2] + [Ln3] + [ln4]

ln1=0,[ln1]=0
one

Is it right to write LN2 as ln1 / 2

At first glance, they are from Jiangxi. Ha ha, I think they are right

What is 1 / 2 (ln9-ln1) equal to

ln9-ln1=ln(9/1)=ln9