Given the piecewise function f (x) = 1, X is greater than or equal to 0; - 1, X

Given the piecewise function f (x) = 1, X is greater than or equal to 0; - 1, X

Discussion on piecewise function when x

lg5*lg20=?
Why is lg5 * LG20 = lg5 (1 + LG2) = lg5 + lg5lg2?
It's clear

lg5*lg20=lg5(lg10+lg2)=lg5(1+lg2)=lg5+lg5lg2
LG 10 = 1

Calculate the following formula (1) (2lg2 + Lg3) / (1 + 1 / 2lg0.36 + 1 / 3lg8) (2) lg5 ^ 2 + 2 / 3lg8 + lg5 * LG20 + (LG2) ^ 2

Original formula = LG12 / (LG10 + lg0.6 + Lg3) = LG12 / LG18 = log18 (12) original formula = 2lg5 + 2lg2 + lg5 * (2lg2 + lg5) + (LG2) & sup2; = (lg5) & sup2; + 2lg5lg2 + (LG2) & sup2; + 2 (lg5 + LG2) = (lg5 + LG2) & sup2; + 2lg10 = (LG10) & sup2; + 2 = 1 + 2 = 3

lg20+lg5=______ ．

The original formula = lg5 + (lg5 + 2lg2) = 2 (lg5 + LG2) = 2lg10 = 2, so the answer is: 2

The calculation of LG20 times lg5
Why is LG20 times lg5 equal to (1 + LG2) * (1-lg2)

lg20*lg5
=[lg(2*10)]*[lg(10/2)]
=(lg2+lg10)*(lg10-lg2)
=(lg2+1)*(1-lg2)
=(1+lg2)*(1-lg2)

Given the logarithm of X, find X. LG x = LG a + LG B
What kind of format?
I didn't even say how to write it. Can I just hand it in like this?

LGx=LGa+LGb=LG(ab),
Since LG function is monotonically increasing, x = ab

Let a > 0, a ≠ 1, x, y satisfy that log is logarithm of base x with a + 3log is logarithm of base a with X - log is logarithm of base y with x = 3
(1) Log is represented by the logarithm of X with a as the base, and log is represented by Y with a as the base
(2) When x takes what value, log takes a as the base and Y gets the minimum value

1 log log with a as the base x + 3 log with X as the base a - log with X as the base y = 3. It can be changed into log with a as the base x + 3 / log with a as the base X - log with a as the base Y / log with a as the base x = 3, that is, the square of log with a as the base x + 3-log with a as the base y logarithm = 3 log with a as the base X

A high school mathematics problem log2, see the supplement
Given log2 [Log1 / 2 (log2 ^ x)] = log3 [Log1 / 3 (log3 ^ y)] = log5 [Log1 / 5 (log5 ^ z)] = 0, try to compare the size of X, y, Z?

Log2 [Log1 / 2 (log2x)] = 0 = log2 (1) Log1 / 2 (log2x) = 1 = Log1 / 2 (1 / 2) log2 (x) = 1 / 2x = 2 ^ (1 / 2) log3 [Log1 / 3 (log3y)] = 0log1 / 3 (log3y) = 1log3 (y) = 1 / 3Y = 3 ^ (1 / 3) x ^ 6 = 2 ^ 3 = 8y ^ 6 = 3 ^ 2 = 9, so Y > xlog5 [Log1 / 5 (log5z)] = Log1 / 5 (log5z) = 1LOG

A mathematical logarithm expansion problem in senior one
How many digits does 2 ^ 60 have?
Ask: simple train of thought and answer process, thank you!

1) The numbers we use are all decimal notation, and each digit of each digit is composed of the product of the number and the weight of 10. Take 6789 as an example, the 6000 represented by the thousand digit 6 is composed of the product of the number 6 and 10 ^ 3; the numbers in other digits are the same
2) Looking back at the problem, ask how many digits there are in 2 ^ 60, it can be understood as seeking the highest weight + 1; for example, any number between 1000 and 9999, such as 6789, has 4 digits, its highest position represents the number 6, and the weight is 10 ^ 3; take the logarithm lg6789 of the number with 10 as the bottom, and get 3.83, the integer is the index 3 of the weight, and the decimal is the logarithm LG6 of the number 6
3) It's easy to figure out the truth. Find LG2 ^ 60 = 18.06 and round it up to + 1, 19 digits

A math problem of senior one about logarithm operation
LGA + LGB = 2lg (a-2b), find the value of a / b