What is the formula for changing bottom?

What is the formula for changing bottom?

Logan = logcn / logca, where n > O, a > O, C > O, and a is not equal to 1, C is not equal to 1
The following equations are solved by factorization
3x²-12x=-12
3x²-12x=-12
x²-4x+4 =0
﹙ X-2﹚²=0
∴X1=X2=2
If X & # 178; - 4x + 4 = 0, then the square of (X-2) = 0
X=2
x^2-4x+4=0
(x-2)(x-2)=0
x1=2 x2=2
How to solve 3x-210 = x + 30
3X-210=X+30
Move the right side to the left, and the positive sign changes to the negative sign
3X-210-X-30=0
That is 2x-240 = 0
X=120
transposition
Move everything on the right to the left
We get 3x-210-x-30 = 0
Solution x = 120
A problem solved by changing the bottom formula
In RT △ ABC, C is the hypotenuse, a and B are the right angle sides
Verification: logC + B (a) + logc-b (a) = 2logc + B (a) * logc-b (a)
(Note: C + B, C-B are all bottom, and "(a)" is true)
C & sup2; = A & sup2; + B & sup2; a & sup2; = (c + b) (C-B) take the logarithm 2lga = LG (c + b) + LG (C-B) left = LGA / LG (c + b) + LGA / LG (C-B) general partition = LGA {[LG (C-B) + LG (c + b)]} / [LG (C-B) * LG (C-B)] = LGA * 2lga / [LG (C-B) * LG (C-B)] = 2 [LGA / LG (C-B)] * [LGA / LG (C-B)] = 2lo
Bottom change:
Left = loga (a) / loga (c + b) + loga (a) / loga (C-B)
=(loga(c+b)+loga(c-b))/loga(c+b)loga(c-b)
=Loga (C square-b Square) / loga (c + b) loga (C-B)
=Loga (a square) / loga (c + b) loga (C-B)
=2/loga(c+b)loga(c-b)
Right = 2 * log a (a) / log a (c + b) * log a (a) / log a (C-B) = 2 / log a (c + b) log a (C-B)
So left = right
Get proof
If the square of 4x + the square of 5x + K has a factorization of 4x-3, then the value of K is
(4x-3)(x+2) k=-6
The solution process of 3x + x = 30
3X+X=30
4X=30
X=30÷4
X=7.5
3x+x=30
4x=30
x=30/4
x=7.5
3X+X=30
4X=30
X=30/4=7.5
3x+x=30
4x=30
x=7.5
Thank you for your adoption``
Solution:
The result of merging similar items is: 4x = 30
Multiply both sides by 1 / 4 to get x = 7.5
Two topics on the formula of changing bottom
Use the bottom changing formula to solve the values of the following formulas:
① log（2）25 × log（3）4 × log（5）9
②（lg2)2+lg2+lg50+lg25
Please master help answer, give the answer process, I am very grateful
Original formula = (LG2 / LG25) * (Lg3 / LG4) * (lg5 / lg9)
=(lg2/lg5^2)*(lg3/lg2^2)*(lg5/lg3^2)
=1/(lg5*lg2*lg3)=1/(lg10)=1
If a, the square of the equation x minus 3x minus the two roots of 5 = 0, find the square of α plus the square of 2 β minus 3 β
a. Two roots of β equation x ^ 2-3x-5 = 0
So a + β = 3
aβ=-5
So α ^ 2 + 2 β ^ 2-3 β
=α^2+2β^2-β(a+β)
=α^2+β^2-aβ
=(a+β)^2-3aβ
=9+15=24
a. Two roots of β equation x minus 3x minus 5 = 0
That is, α & sup2; = 3 α + 5
β & sup2; - 3 β - 5 = 0, that is β & sup2; = 3 β + 5
And α + β = 3
The square of α plus the square of 2 β minus the value of 3 β
=3α+5+2（3β+5）-3β
=3（α+β）+15
=9+15
=24
x^2-3x-5=0
α+β=3
α^2 + 2β^2-3β = [α^2 -3α-5] +2[β^2-3β-5] + 3α+5+6β+10 -3β
=0+0+3[α+β]+15
=24
(x+17)÷[(3x+45)÷10]=2
（x+17)÷[(3x+45)÷10]=2
x+17=2(3x+45)÷10
x+17=(3x+45)÷5
5x+17*5=3x+45
5x-3x=45-85
2x=-40
x=-20
(x+17)÷[(3x+45)÷10]=2
x+17=2×[(3x+45)÷10]
x+17=1/5(3x+45)
x+17=3x/5+15
x-3x/5=15-17
2x/5=-2
x=-2*5/2
x=-5
(x+17)÷[(3x+45)÷10]=2
(x+17)÷(3x+45)×10=2
10(x+17)=2(3x+45)
10x+170=6x+90
10x-6x=90-170
4x=-80
x=-20
The application of bottom formula!
Excuse me, what is the significance of the formula in calculation? What is its use? Why do we need it? And do we need common logarithms at the same time?
The formula of changing base can be expressed as (‖ below, () is the base number, and [] is the true number ‖)):
log(a)[b]={log(p)[]b}/{log(p)[a]}
The function is to change the logarithm into one with the same base, which is convenient for logarithm operation. The base usually has nothing to do with the operation, so it usually takes 10 or E as the base