Why the base of logarithm a must be greater than 0 and less than 0?

1. Logarithms come from exponents
Log (a) n = B is equivalent to a ^ B = n
2. When discussing the index, when the base a is negative, the situation is more complicated, so it is not easy to summarize the general things
In order to facilitate students to understand and accept, it is generally stipulated that a > 0, a ≠ 1
But it's not that a < 0 can't be discussed, it's just that there is no special discussion in middle school
3. Similarly, when discussing logarithm, when the base a is negative, the situation is more complex, and it is not easy to generalize the general things. In order to be consistent with the discussion of exponential function, and for the convenience of students' understanding and acceptance, it is generally stipulated that a > 0, a ≠ 1
But it's not that a < 0 can't be discussed, it's just that there is no special discussion in middle school

If the logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | logarithm of | loga

Logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm of logarithm

loga(2)>logb(2),
According to the formula, 1 / log2 (a) > 1 / log2 (b), (*)
1) If log2 (a) > 0, log2 (b) 1 > b > 0;
2) If log2 (a) > 0, log2 (b) > 0, log2 (a) a > 1 is obtained from (*);
3) If log2 (a)

We know the logarithm with m as the base 3

A = log_ m(3),b=log_ n(3)
According to the definition of logarithm: m ^ a = 3, n ^ B = 3
m=3^(1/a),n=3^(1/b)
From the title meaning A1 / B, M > n

Compare the size of each logarithm below and give reasons
(1) - 4 and 3. (2) 0 and -0.005. (3) - three fifths and - four fifths

-4-0.005
-Three fifths > - four fifths

Compare the following logarithms and give reasons
-6 out of 7 and - 7 out of 8 | - (+ 1 out of 2) | - and - (- 1)

Solution-6 / 7 - (- 7 / 8)
=-6/7+7/8
=-48/56+49/56
=-1/56＜0
That is - 6 / 7 ＜ (- 7 / 8)
|-(+ 1 / 2) | = 1 / 2 < 1 = - (- 1)
That is | - (+ 1 / 2) | - (- 1)

(logarithm of base 5 with 2 + logarithm of base 0.2 with 4) times (logarithm of base 2 with 5 + logarithm of base 0.5 with 25)

The logarithm of base five with two as the base + the logarithm of base 0.2 with four as the base) multiplied by (logarithm of base two with five as the base + logarithm of base 0.5 with twenty five as the base) = (log25 + log40.2) (log52 + log250.5) = (log25 + log20.2 / log24) (log52 + log50.5 / log525) = (log25 + log20.2 / log22) = (log

How to calculate the logarithm of 0.5 based on 0.84

Using the same base formula: A is the base, (loga) B = (log10) B / (log10) a
Then (log0.84) 0.5 = (log10) 0.5 / (log10) 0.84 = (log10) 50 / (log10) 84

(lg√2+lg3-lg√10)÷lg1.8

(lg√2+lg3-lg√10)÷lg1.8
=(1/2)(lg2+lg9-lg10)÷lg1.8
=(1/2)lg1.8÷lg1.8
=1/2

Can we use logarithm?
Given that the 11th power of X is equal to 1.35, find X
The former friend answered the value of X, but did not answer whether or how to use logarithm to solve this problem; the latter friend listed the logarithm solution, but there was no result. I hope which one can answer completely, the more detailed the better. I really want to find out this problem.

Is halo a power exponent? X = 11sqrt (1.35) = 1.027657803533499

Why the base of logarithm a must be greater than 0 and less than 0?