Y = log logarithm with 2 as the base + log with X as the base, and the range of logarithm of 2x is

Y = log logarithm with 2 as the base + log with X as the base, and the range of logarithm of 2x is

y=log2 x+logx 2x
=log2 x+log2 2x/log2 x
=log2 x+(log2 2+log2 x)/log2 x
=log2 x+1/log2 x+1
Here we use the basic inequality
|log2 x+1/log2 x|≥2
So log2 x + 1 / log2 x ≥ 2 or ≤ - 2

It is known that log takes 18 as the base, the logarithm of 9 = a, and the power of 18 = 5. The formula containing a and B is used to express log, with 36 as the base and 45 as the logarithm

A = log18 9, B = log18 5. Therefore, log3645 = (log18 45 / log18 36) is used to change the base of the formula, and log18 45 = log18 (5 * 9) = (log18 5) + (log18 9) = a + B log18 36 = log18 (4 * 9) = (log18 9) + [log18 (9-5)] = a + A / b. therefore, log3645 = (a + b) / (a + A / b) = (a

The base of the known logarithm is log: B, which is the base of log

Log (18) [5]: denotes the logarithm of 5 with 18 as the base. If the B power of 18 = 2, then: log (18) [2] = B, then: log (45) [36] [base changing formula] = [log (18) [45]] / [log (18) [36] = [log (18) [5] + log (18) [9] / [1 + log (18) [2]] = [log (18) [5] + 1 - log (18) [2]] / [

If the a power of 2 = 3, log is represented by a with the base of 12 and the logarithm of 18

Log takes the logarithm of base 12 and 18 into
Log2 (18) divided by log2 (12)
Log2 (18) = log2 (3? Times 2) = 2log2 (3) + log2 (2)
Log2 (12) = log2 (2? Times 3) = log2 (3) + log2 (2?)
From the a power of 2 = 3
Log2 (3) = a
The original formula = (2a + 1) / (a + 2)

It is known that the logarithm of log with base 18 as base 9 is a, and the number B of 18 is 5. Try to use a and B to represent the logarithm of log with base 36 as base 45 Urgent,

18^a=9 18^b=5
a=log18 9
b=log18 5
log36 (45)=log18(45)/log18(36)
=(log18 5+log18 9)/(log18 18 +log18 2)
=(a+b)/(2-a)

It is known that log takes 18 as the base, the logarithm of 9 = a, and the power of 18 = 5. The formula containing a and B is used to express log. Log takes 36 as the base and 5 as the logarithm

From known
a=log18(9)=log2(9)/log2(18)=2log2(3)/[1+2log2(3)]
b=log18(5)=log2(5)/log2(18)=log2(5)/[1+2log2(3)]
From this solution, log2 (3) = A / [2 (1-A)], log2 (5) = B / (1-A),
Therefore, according to the logarithm change bottom formula
log36(5)=log2(5)/log2(36)=log2(5)/[2+2log2(3)]=b/(2+a).

The logarithm of log base 2 is a, the logarithm of log base 3 is B, and the logarithm of log base 42 is 56?

a=lg3/lg2
lg3=alg2
b=lg7/lg3
lg7=blg3=ablg2
log42(56)
=lg56/lg42
=lg(2^3*7)/lg(2*3*7)
=(3lg2+lg7)/(lg2+lg3+lg7)
=(3lg2+ablg2)/(lg2+alg2+ablg2)
=(3+ab)/(1+a+ab)

Given that log base 2 logarithm of 3 = a, log base 3 logarithm = B, find log base 14 base 56 logarithm

A:
a=log2(3),b=log3(7)=log2(7)/log2(3)
So: ab = log2 (7)
log14(56)
=log2(7*8)/log2(14)
=[log2(7)+3]/[log2(7)+1]
=(ab+3)/(ab+1)
This question is to examine the logarithmic function of the bottom formula

What is the log of 3 based on 4 and the logarithm of 3? RT, Really can't do it? Is it not clear? To the power of 3 Is that clear this time?

Upstairs, I'm sure to tell you, this is not the simplest, I can simplify it, but I don't want to type those mathematical symbols now, and it's useless to play them. In a short time, I'm not sure that I can get the value, or I'll calculate it

How much is the [- (log base on 3, logarithm of 2)] of 3? How many powers of root 3 equal to 1? RT

The [- (log base on 3, logarithm of 2) power]
=3 [(log based on 3, logarithm of 2 to the power of - 1)]
=3 [(log base 3, logarithm of 1 / 2) power]
=1/2
When a ≠ 0
A to the power of 0 = 1
So the root 3 to the power of 0 is 1