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Operation of logarithm (1) Log to the 1 / 100 power of 10 =? (2) Log to the 18th power of 3 - log to the 2nd power of 3 =? (3) What is the value of log with base 8 and true number 9 compared with log with base 2 and true number 3? (4) Log takes 2 as the base true number and x = log takes 8 as the base true number and X, then the value of X is?
1. Log to the 1 / 100 power of 10 = log to the 1-log of 10 = 0-2 = - 2
2. Log to the 18th power of 3 - log to the 2nd power of 3 = log to the 9th power of 3 = 2
3. Log with 8 as the base, true number is 9 = (2 / 3) log with 2 as the base, true number is 3
So the answer is 2:3
4. Log takes 2 as the base, true number is x = log takes 8 as the base, true number is X
Log takes 2 as the base, the true number is X - (1 / 3) log takes 2 as the base, the true number is x = 0
Log takes 2 as the base, the true number is x-log takes 2 as the base, the true number is x to the third power = 0
Log is based on 2, the real number is (1 / (x square)) = 0
-Log is based on 2, the true number is x square = 0
Log takes 2 as the base, true number is x square = 0 = log takes 2 as the base, true number is 1
x^2=1
X = 1 or x = - 1 (rounding off)
So x = 1
2lg5
Five is the real number
Because it's LG, the base number is 10
I would like to ask if the base number 10 and the front two times can be approximately divided
Get log5 (5)
The base number becomes 5
Isn't the formula equal to 1
Mathematics is also very strong for your part
2lg5=lg(5*5)=lg25
Mathematical exponential logarithm
2 ^ 6A = 3 ^ 3B = 6 ^ 2C, find 1 / 6A + 1 / 3b-1 / 2C? Answer zero
Let 2 ^ 6A = 3 ^ 3B = 6 ^ 2C = a6a = log ﹥ 8322; a3b = log ﹥ 8323; A2C = log6, a 1 / 6A + 1 / 3b-1 / 2C = 1 / log ﹥ 8322; a + 1 / log ﹥ 8323; A-1 / log6, a = log (a) 2 + log (a) 3-log (a) 6 = log (a) (2 × 3 △ 6) = log (a) (1) = 0
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