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Given log (12) 27 = a, find the value of LG (2) / LG (3)
lg(27)/lg(12)=a
lg(2 × 3)/lg(3)=1/a
(2lg2+lg3)/3lg3=1/a
(2lg2)/(3lg3)=(1/a)－1/3
(lg2)/(lg3)=(3/2a)－1/2
What is the difference between LG and log in a logarithmic function? How did log become LG?
LG is the log with a base of 10, and the log can take other bases greater than 0
Known α，β Is two of equation lg2x-lgx-2 = 0, then log αβ+ log βα The value of is () A. 3 B. 2 C. -5 two D. −3 two
∵ α，β Are two of equation lg2x-lgx-2 = 0,
Solve the equation lg2x-lgx-2 = 0 to obtain:
Lgx = - 1 or lgx = 2,
The solution is x = 1
10, or x = 100,
∴ α＝ one
10， β= 100, or α＝ 100， β＝ one
10，
log αβ+ log βα= log1
10100+log1001
10=-2+（-1
2）=-5
2．
Therefore: C
(log(3)2+lg(9)2)(log(4)3+log(8)3) The numbers in parentheses are the base numbers, and the rest are real numbers, which should be calculated by the bottom changing formula
(log (3) 2 + LG (9) 2) (log (4) 3 + log (8) 3) = (LN2 / Ln3 + LN2 / ln9) (Ln3 / ln4 + Ln3 / ln8) = [LN2 / Ln3 + LN2 / (2ln3)]
Please tell me all the formulas in the log,
1. Log (c) (a * b) = log (c) a + log (c) B - equivalent to multiplying the power of the base, the base remains unchanged "exponential addition" log (c) (A / b) = log (c) a / log (c) B - equivalent to dividing the power of the base, the base remains unchanged "exponential subtraction" 2. Log (c) (a ^ n) = n * log (c) a - equivalent to the power of the power, the base
It is known that a > 1, b > 1, C > 1, and the logarithm of log with a as the base C multiplied by the logarithm of log with B as the base C is equal to 4. It is proved that ab > = C Brother, hurry up
Because logac = 1 / logca and Logbc = 1 / logcb, the original formula is logca * logcb = 1 / 4,
That is, the root sign (logca * logcb) = 1 / 2. According to the inequality: 2 root signs ab
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