What are the operational properties of logarithm

What are the operational properties of logarithm

Basic properties
If a > 0, and a ≠ 1, M > 0, n > 0, then:
1、a^log(a)(b)=b 　　
2、log(a)(a)=1 　　
3、log(a)(MN)=log(a)(M)+log(a)(N)　　
4、log(a)(M÷N)=log(a)(M)-log(a)(N) 　　
5、log(a)(M^n)=nlog(a)(M) 　　
6、log(a)[M^（1/n）]=log(a)(M)/n 　　
(Note: ^ are superscripts, for example: A ^ 1 is a)
7. The formula for changing the bottom is as follows:
log(a)(N)=log(b){N}÷log(b){a}
8、log(a){b}=1/log(b){a}

Evaluation: lg5 & # 178; + LG2 * LG50 + (LG2) &# 178;
The answer is: 2lg5 + LG2 * LG (5 * 10) + (LG2) & # 178; ① = 2lg5 + LG2 * lg5 + LG2 + (LG2) & # 178; ② = 2lg5 + LG2 * (lg5 + LG2) + LG2 ③
Step 2: How did you get it? How did it become LG2 * lg5 + LG2?

lg5²+lg2*lg50+(lg2)²
=2lg5+lg2*lg(5*10)+(lg2)²
=2lg5+lg2*(lg5+lg10)+(lg2)²
=2lg5+lg2*(lg5+1)+(lg2)²
=2lg5+lg2*lg5+lg2+(lg2)²
=2lg5+lg2+[lg2*lg5+(lg2)²]
=2lg5+lg2+lg2*(lg5+lg2)
=2lg5+lg2+lg2
=2(lg5+lg2)
=2

The operation of logarithm in senior one
1/2 lg32/49-4/3 lg√8+lg√245
process

Original formula = LG (32 / 49) ^ 1 / 2 - LG8 ^ 1 / 2 * 4 / 3 + lg7 √ 5
= lg√32-lg√49-lg4+lg7+lg√5
=lg4√2-lg7-lg4+lg7+lg√5
=lg√2+lg√5
=lg√10
=1/2

Log12 27 = a, find log6 16?
Log147 = a, log145 = B, find log3528?
Lg5 = a, log26 = B, find log32, log40.12?
Log6 15 = a, log5 18 = B, find log30 50?

Do a problem for you to demonstrate, I believe you can draw inferences from one example. First problem: a = ln27 / ln12 (change logarithm of the same base, usually e as the base) = 3ln3 / (2ln2 + Ln3) (decompose into prime number) then get LN2 / Ln3 = (3-A) / (2a) and then find log616, the same method: log616 = ln16 / ln6 = 4ln2 / (LN2 + Ln3)

Question 1 (LG radical 2 + Lg3 LG radical 10) / lg1.8
In question 2, we know that the logarithm with 6 as the base, 7 = a, the logarithm with 3 as the base, 4 is B, find the logarithm with 12 as the base, 7?

(LG radical 2 + Lg3 LG radical 10) / lg1.8
=lg(√2*√9/√10)/lg1.8
=lg√1.8/lg1.8
=(1/2)*lg1.8/lg1.8
=1/2
log6(7)=a
lg7/(lg2+lg3)=a
lg7=alg2+alg3
log3(4)=b
2lg2/lg3=b
lg2=(b/2)lg3
So lg7 = a (B / 2) Lg3 + alg3 = (a + AB / 2) Lg3
log12(7)
=lg7/(2lg2+lg3)
=(a+ab/2)lg3/[2*(b/2)lg3+lg3]
=(a+ab/2)/[2*(b/2)+1]
=(2a+ab)/(2b+2)

Let LG2 = A and Lg3 = b be known, and find the values of the following formulas:
(1) Log3 4 (the first 3 is the base of logarithm and the last 4 is true)
(2) Log2 12

log3(4)=lg4/lg3
=2lg2/lg3
=2a/b
log2(12)
=lg12/lg2
=lg(3×2²)/a
=(lg3+lg2²)/a
=(b+2lg2)/a
=(b+2a)/a

Given INX + iny = 4, find the minimum value of 1 / x + 1 / y

Inx+Iny=4
x>0 y>0
ln(xy)=4
xy=e^4
1/x+1/y
>=2/√xy=2/e^2
If and only if:
When 1 / x = 1 / y = 1 / e ^ 2, the equal sign holds
So: min (1 / x + 1 / y) = 2 / e ^ 2

Logarithm and logarithm operation in senior one
Log2 (25) * log3 (4) * log5 (9) ()

Log2 25 * log3 4 * log5 9 = log2 5 ^ 2 * log3 2 ^ 2 * log5 3 ^ 2 = 2 * log2 5 * 2 * log3 2 * log5 3 = 8 * (log2 5) * (log2 2 2) / (log2 3) * (log2 3) / (log2 5) = 8 * log2 2 = 8 * 1 = 8 or log2 (25) * log3 (4) * log5 (9) = log25 / log2 * log4 / log3 * log

Prove logarithmic inequality
Let 0 < a < 1, x < 0
ln[√（x²+1）+x]＜x（a^x-1)/[(a^x+1)log√(x²+1)-x]

Because x

Given that the image of F (x) = log a [(1-mx) / (x-1)] is symmetric about the origin, the value of M is obtained
It's a bit hard to understand. A is the base number and [(1-mx) / (x-1)] is the true number

f(-x)=-f(x)
(1-mx)/(x-1)=((1+mx)/-x-1)^-1
1-m^2 * x^2=1-x^2
M = + - 1, when m = 1, f (x) has no significant rounding off, so m = - 1