Given that the function f (x) = x ^ 2 + X * LG (a + 2) + LGB satisfies f (- 1) = - 2, for all real numbers x, f (x) ≥ 2x, find the value of real number a * B

Given that the function f (x) = x ^ 2 + X * LG (a + 2) + LGB satisfies f (- 1) = - 2, for all real numbers x, f (x) ≥ 2x, find the value of real number a * B

From F (- 1) = - 2, 1-lg (a + 2) + LGB = - 2 (1)
Because for all real numbers x, f (x) > = 2x,
Therefore, x ^ 2 + X (LG (a + 2) - 2) + LGB > = 0 holds, that is, the equation x ^ 2 + X (LG (a + 2) - 2) + LGB = 0 has at most one root
Therefore, △ = [LG (a + 2) - 2] ^ 2-4lgb

Given that log is based on 18, 9 = a, 18 ^ B = 5, find log is based on 36, 45

log18 9=a
log18 5=b
Log36 45 = log18 45 / log18 36 = log18 (5 times 9) / log18 (2 times 18) = (log18 5 + log18 9) / (log18 2 + log18 18) = (a + b) / (1 + log18 2)
Where log18 2 = log18 (18 / 9) = log18-log18 9 = 1-A
So the original formula = (a + b) / (2-A)

The derivation of three operational properties of logarithm

The property and derivation of logarithm
Use ^ to express the power and log (a) (b) to express the logarithm of B with a as the base
*Is the multiplication sign, / is the division sign
Definition formula:
If a ^ n = B (a > 0 and a ≠ 1)
Then n = log (a) (b)
Basic properties:
1.a^(log(a)(b))=b
2.log(a)(MN)=log(a)(M)+log(a)(N);
3.log(a)(M/N)=log(a)(M)-log(a)(N);
4.log(a)(M^n)=nlog(a)(M)
deduction
1. You don't need to push this. You can get it directly from the definition (take [n = log (a) (b)] in the definition into a ^ n = b)
two
MN=M*N
From the basic property 1 (replace m and N)
a^[log(a)(MN)] = a^[log(a)(M)] * a^[log(a)(N)]
By the nature of index
a^[log(a)(MN)] = a^{[log(a)(M)] + [log(a)(N)]}
And because exponential function is monotone function, so
log(a)(MN) = log(a)(M) + log(a)(N)
3. Similar to 2
MN=M/N
From the basic property 1 (replace m and N)
a^[log(a)(M/N)] = a^[log(a)(M)] / a^[log(a)(N)]
By the nature of index
a^[log(a)(M/N)] = a^{[log(a)(M)] - [log(a)(N)]}
And because exponential function is monotone function, so
log(a)(M/N) = log(a)(M) - log(a)(N)
4. Similar to 2
M^n=M^n
From basic property 1 (replace m)
a^[log(a)(M^n)] = {a^[log(a)(M)]}^n
By the nature of index
a^[log(a)(M^n)] = a^{[log(a)(M)]*n}
And because exponential function is monotone function, so
log(a)(M^n)=nlog(a)(M)
Other properties:
Property 1: bottom changing formula
log(a)(N)=log(b)(N) / log(b)(a)
The derivation is as follows
N = a^[log(a)(N)]
a = b^[log(b)(a)]
It can be obtained by synthesizing the two formulas
N = {b^[log(b)(a)]}^[log(a)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
And because n = B ^ [log (b) (n)]
therefore
b^[log(b)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
therefore
Log (b) (n) = [log (a) (n)]
So log (a) (n) = log (b) (n) / log (b) (a)
Property 2: (I don't know the name)
log(a^n)(b^m)=m/n*[log(a)(b)]
The derivation is as follows
According to the formula [LNX is log (E) (x), e is called the base of natural logarithm]
log(a^n)(b^m)=ln(a^n) / ln(b^n)
From the basic properties 4
log(a^n)(b^m) = [n*ln(a)] / [m*ln(b)] = (m/n)*{[ln(a)] / [ln(b)]}
And then by changing the bottom formula
log(a^n)(b^m)=m/n*[log(a)(b)]
--------------------------------------------(end of properties and derivation)
Formula 3:
log(a)(b)=1/log(b)(a)
The proof is as follows:
From the formula log (a) (b) = log (b) (b) / log (b) (a) --- take the logarithm with B as the base, log (b) (b) = 1
=1/log(b)(a)
It can also be deformed into:
log(a)(b)*log(b)(a)=1

How to deduce the property of logarithm operation 3

Which nature?
Using definition, strict reasoning
log(a)(M^n)=nlog(a)(M)
Let a ^ n = M
so
n=loga(M)
loga(M^n)=loga(a^n^2)=n^2=nloga(M^n)
Get proof

The nature of logarithm operation is the derivation of formulas (all)
In particular, logam n (index) = nlogam

loga（MN）=logaM+logaN
prove:
Let logam = P, Logan = Q. The definition of logarithm can be written as M = AP, n = aq
M·N=ap·aq=ap+q,
therefore
loga（M·N）=p+q=logaM+logaN．
Namely
loga（MN）=logaM+logaN．
Every logarithm has meaning, that is m > 0, n > 0; a > 0 and a ≠ 1
Division is the same. Thank you
attach
It is proved that logam n (exponent) = nlogam
If logam = x, Logan = y, then a ^ x = m, a ^ y = n ｜ Mn = a ^ XA ^ y = a ^ (x + y), then x + y = loga (MN), that is, logam + Logan = logamn, let logam = x, that is, a ^ x = m, then (a ^ x) n = m ^ n, that is, a ^ (NX) = m ^ n ｜ loga ^ m (^ n) = NX = nlogam
Get proof

The property and derivation of logarithm

Does it refer to its operational properties?
Logarithm of product = sum of logarithms;
Logarithm of quotient = difference of logarithm;
Logarithm of power = exponent × logarithm

Proof formula: logamn = nlogam (n ∈ R)

Prove that logam ^ n = nlogam (n is a real number)
Both a's are base numbers

Logam ^ n = loga (m * m * m *... * m) total N M
Log a (m * m * m *... * m) = log a m + log a m +... + log a m, n log a m = nllog a m
∴logam^n=nlogaM

Why does the limit algorithm of power exponential function f (x) ^ g (x) stipulate that f (x) > 0 and not equal to 1? F (x) is greater than 0 because it is derived from e ^ LNF (x) ^ g (x)
Is it not equal to 1 because it is meaningless to discuss this (founder is the same)?

If you want to investigate, it is true that f (x) = 1 is an ordinary case, which is not worth studying. But if the condition of the law Lim f (x)! = 1, it is necessary, because Lim g (x) = oo

Given the function f (x) = 1, X is less than or equal to 0. 2 F (x-1) + 1, x > 0. The following order of the equation f (x) = x is from small to large
Then the general term formula of the sequence is

When x0, f (x) = x, (x is a positive integer)
So the general formula listed in this paper is an = n-1