log5(3)+log5(1/3)=? How to calculate

log5(3)+log5(1/3)=? How to calculate

log5(3)+log5(1/3)=log5(3*1/3)=log5(1)=0

Log5 (1 / 3) x log36 x log6a = 2, find a

Log5 (1 / 3) x log36 x log6a = 2 by using the bottom changing formula, [LG (1 / 3) / lg5] x [LG6 / Lg3] x [LGA / LG6] = 2 LG6 reduced, [- Lg3 / lg5] x [LGA / Lg3] = 2 Lg3 reduced, LGA / lg5 = - 2 log5a = - 2 a = 1 / 25

log5(125) =3
The third power of 5 = 125
How about this three

log5(5)+log5(5)+log5(5)=log5(5*5*5)=log5(125)

Log (4) log (5) M = 35

log(5)35=m
lg35/lg5=m
(lg5+lg7)=mlg5
lg7=(m-1)lg5
log(7)1.4=lg1.4/lg7=(lg(7/5))/lg7
=(lg7-lg5)/lg7
=1-lg5/lg7
=1-1/(m-1)
=(m-2)/(m-1)

Log function 2log5 (15) - log5 (18) + log5 (7)

log5(225*7/18)=log5(175/2)

Who is bigger in log5 ^ 4 and 2log5 ^ 3? How to compare them?

Solution 2log5 Λ 3 = log5 Λ 3 & #178; = log5 Λ 9
Because the base is the same and greater than 1 and y = log5, Λ x is an increasing function
9 ＞ 4, so log5 ∧ 9 ＞ log5 ∧ 4
That is log5 ∧ 4 < 2log5 ∧ 3

Simplify (a + 1) (A-2) / (a + 2) (A-2) + 3

=(a+1)(a-2)/(a²-4+3)
=(a+1)(a-2)/(a²-1)
=(a+1)(a-2)/(a+1)(a-1)
=(a-2)/(a-1)

Simplification (a ^ 3 + 1 / A ^ 3) (a ^ 3-1 / A ^ 3) / [(a ^ 4 + 1 / A ^ 4 + 1) (A-1 / a)]

(a^3+1/a^3)(a^3-1/a^3)/[(a^4+1/a^4+1)(a-1/a)]
=(a^6-1/a^6)/[(a^4+1/a^4+1)(a-1/a)]
=[(a^2)^3-1/(a^2)^3]/[(a^4+1/a^4+1)(a-1/a)]
=[(a^2-1/a^2)*(a^4+1/a^4+1)]/[(a^4+1/a^4+1)(a-1/a)]
=(a^2-1/a^2)/(a-1/a)
=a+1/a

When 3 < a < 4, simplify | A-3 | + | A-4|=______ ．

∵ 3 ＜ a ＜ 4 | A-3 | = A-3, | A-4 | = 4-A, | original formula = | A-3 | + | A-4 | = A-3 + (4-A) = 1

Simplification; (a + b) ^ 2-3 / 2 (a + b) - 1 / 4 (a + b) ^ 2 + (- 2) ^ 3 (a + b)

Original formula = [(a + b) ^ 2-1 / 4 (a + b) ^ 2] + [- 3 / 2 (a + b) + (- 2) ^ 3 (a + b)]
=3 / 4 (a + b) ^ 2-19 / 2 (a + b)