Solve the equation 4x + 66.5 = 6x We need to check the calculation

Solve the equation 4x + 66.5 = 6x We need to check the calculation

X=33.25
66.5=6X-4X
66.5=2X
X=66.5/2
X=33.25
How to understand the exchange formula of logarithm
Let x = a ^ m, a = B ^ n, then x = (b ^ n) ^ m = B ^ (MN) ① The logarithm with a as the base is log (a, x) = M ② The logarithm with B as the base is log (B, x) = Mn ③ (2) log (B, x) / log (a, x)
If x2 + 4x-4 = 0, then 3x2 + 12x-5=______ .
∵ x2 + 4x-5 = 0, ∵ x2 + 4x = 5, ∵ 3x2 + 12x-5 = 3 (x2 + 4x) - 5 = 3 × 4-5 = 7
5 (4x + 20) = 4 (6x-20) + 2
20x＋100=24x-80＋2
20x-24x=-100-80＋2
-4x=-178
4x=178
x=44.5
20x+100=24x-80+2
4x=178
x=44.5
On the deduction of the formula of changing base of mathematical logarithm
Given that log (2) (3) = a, log (3 (7) = B, use a and B to denote log (42) (56)
Because log (2) (3) = a, then 1 / a = log (3) (2), and ∵ log (3) (7) = B,
∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1
Because I haven't been in touch with it for more than four years... I completely forget how the principle is... Please help me to explain it carefully... Thank you... Especially log (3) (56) / log (3) (42) = log (3) (7) + 3 · log (3) (2) / log (3) (7) + log (3) (2) + 1 = AB + 3 / an + B + 1... I don't know what method is used here... The account I just applied for has less points... Please forgive me···
Remember the formula, ㏒ (a) (b) and ㏒ (b) (a) are reciprocal, log (2) (3) = a, then 1 / a = log (3) (2), which is the reason for this step,
㏒ (a) (b) = ㏒ (c) (b) / ㏒ (c) (a), so ㏒ (42) (56) = ㏒ (3) (56) ㏒ (3) (42)
㏒ (AB) = ㏒ a + ㏒ B, so log (3) (56) = log (3) (7) + log (3) (8)
㏒ a ^ B (B power of a) = B ㏒ a, so ㏒ (3) (8) = 3 ㏒ (3) (2)
The denominator is the same. You're looking at it
(3x^2+9x+7)/(x+1)-(2x^2+4x-3)/(x-1)-(x^3+x+1)/(x^2-1)=
=[(3x^2+9x+7)(x-1)-(2x^2+4x-3)(x+1)-(x^3+x+1)]/(x^2-1)=(4x+5)/(1-x^2)
Solution equation: (4x + 20) / 4 - (6x-20) / 5 = 2
Multiply both sides by 20
5(4X+20)-4(6X-20)=40
20X+100-24X+80=40
24X-20X=100+80-40
4X=140
X=140÷4
X=35
（4X+20）/4-（6X-20）/5 =2
5×（4X+20）-4×（6X-20） =2×20
20X+100-24X+80 =40
4X=100+80-40
4X =140
X=35
(4X+20)/4-(6X-20)/5=2
Multiply both sides of the equation by 20 5 * (4x + 20) - 4 * (6x-20) = 40
20X+100-24X+80=40
24X-20X=100+80-40
4X=140
X=35
How to prove the deduction of the logarithm exchange formula
a^logc b=b^logc a(a>0,b>0,c>0,c≠1)
Let logC (a) = m, logC (b) = n,
Because logC (a) & _; logC (b) = logC (b) & _; logC (a)
So mlogc (b) = nlogc (a)
logc(b)^m=logc(a)^n
b^m=a^n
That is, a ^ log (c) B = B ^ logC (a)
Can you help me? The book doesn't write clearly what the bottom changing formula is, how it is deduced and its inference, so it helps me to answer it. thank you!! The proof process is shown in the figure below: bottom changing formula
What is 4x-8 = 3x + 24
4x—3x=24+8 x=32
Solution equation: 4 (4x + 20) = 5 (6x-20)
4（4x+20）=5（6x-20）
16X+80=30X-100
30X-16X=100+80=180
14X=180
X=90/7