In the right triangle ABC, C is the hypotenuse, a, B is the right side, the proof: log (c + b) ^ A + log (C-B) ^ a = 2log (c + b) ^ log (C-B) ^ a

In the right triangle ABC, C is the hypotenuse, a, B is the right side, the proof: log (c + b) ^ A + log (C-B) ^ a = 2log (c + b) ^ log (C-B) ^ a

Mathematical problems about log
Log147 = B, log145 = B, log3528 is represented by a and B

This problem is actually log147 = a, log145 = B. use a and B to represent log3528, right
log35 28=1/log28 35
log28 35=1/2（log14 7+log14 5）=1/2（a+b）
So log35 28 = 2 / (a + b)

Mathematical problems about log
When 0

First of all:
x=a^loga（x）
Therefore, x ^ loga (x) > A ^ 2 becomes
a^[loga（x）]2>a^2,
And then by 0

How about log (2) 36-log (2) 12

When the base is the same, the two logarithms are subtracted, which is equivalent to the same base and divided exponentially, so it is equal to log (2) (36 / 12)

Log &; (3 &; × 27 &;) is a process

log₃3^4+log₃27^2
=4log₃3+2log₃3^3
=4log₃3+6log₃3
=4+6
=10

When n > 2, prove: logn (n-1) times logn (n 1)

The following are all natural logarithms based on 10
From the mean inequality
lg(n-1)lg(n+1)＜{[lg(n-1)＋lg(n+1)]/2}^2
＜{[lgn^2]/2}^2=lgnlgn
So [LG (n-1) / LGN] [LG (n + 1) / LGN] < 1
That is logn (n-1) logn (n + 1) < 1

O (n * n) and O (n * logn)
rt
Can we be more specific on the first floor,

In the data structure, each algorithm has its time complexity, which is represented by O (). In brackets, the time complexity is calculated by the algorithm, and N * n is of course the square of n
If the solution is n * n + N, then its time complexity is still the same
0 (n * n) because the rule is to take its highest power

Verification: logn (n-1) times logn (n + 1) 1)

Proof: when 10, that is
When [logn (n-1)] [logn (n + 1)] 2, logn (n-1) > 0, logn (n + 1) > 0
So [logn (n-1)] [logn (n + 1)]

When n > 2, prove: logn (n-i) logn (n + 1)

The following are all natural logarithms based on 10
From the mean inequality
lg(n-1)lg(n+1)＜{[lg(n-1)＋lg(n+1)]/2}^2
＜{[lgn^2]/2}^2=lgnlgn
So [LG (n-1) / LGN] [LG (n + 1) / LGN] < 1
That is logn (n-1) logn (n + 1) < 1

We know that n is a natural number greater than 1, and prove that logn (n + 1) > logn + 1 (n + 2)

logn(n+1)=ln(n+1)/ln(n)={ln(n)+ln[(n+1)/n]}/ln(n)=1+ln[(n+1)/n]/ln(n)
Similarly, logn + 1 (n + 2) = 1 + ln [(n + 2) / (n + 1)] / ln (n + 1)
(n+1)/n>(n+2)/(n+1) => ln[(n+1)/n]>ln[(n+2)/(n+1)]
Ln (n) 1 + ln [(n + 2) / (n + 1)] / ln (n + 1)
Then logn (n + 1) > logn + 1 (n + 2)