Logarithmic problems in senior one, logarithmic problems in math in senior one

First of all, let's talk about the idea, 56 = 2*2*2*7, so only log42 2 and log42 7 are required

3=log3 27=log3 12 *log12 27, with log3 2=3 2a-1 2

b=log3 7=log3 42 *log42 7=(1+log3 2+log3 7)*log42 7

Solve log42 7=b (1+b+3 2a-1 2)log3 2=log3 42 *log42 2, solve log42 2 and finally log42 56=3*log42 2+log42 7

Solution: a x=b y=c z=30 w

a=30 (w x), b=30 (w y), c=30 (w z)1 x + 1 dong xin y + 1 z = 1 w nano wheel w x + w y + w z = 1abc = 30 (w x + w charter wheel y + w z) =30 1 = 30

a, b, and c are positive integers.

Or when one of a, b, and c is 1, 1=a x=b y=c z=30 w, w=0, and w is a contradiction to the denominator.

and known by a b c.

a=2,b=3,c=5

Solution: There's a formula like this.

logm^n(a)=1/nlogm

a) (i.e., the logarithm of a at the base of m's nucleus to the nth power = 1 n multiplied.) The opponent of a with m as the base is the number of rocks (Bi Shi Yu) original formula = loglog

log2(7 5) again.

LG2=A,LG2+LG5=LG2*5=1 LG5=1-A, while log2(7)=LG7 LG2=B A,log2(5)=LG5 LG2=(1-A) A; Original = 2 3

log2（7/5）=2/3(log

7)-log2(5))=2/3*[b/a-(1-a)/a]=2(a+b-1)/3a

1 solution: f(x)=log2(x 8)*log1 2(4 x)=(log2(x)-3)(log2(x)-2)=[log2(x)-5 2] 2-1 4, so when in (1 4,8) the value range is [-1 4,,20).

2.Because a=,b=,a=,b=,so.

a b = (1,10), so x a b, then 0 lgx 1, x x 2, so.

lgx) 2 lgx lgx 2, lglgx 0, so there is lglgx (lgx) 2 lgx 2

3.Because f(x)=loga[(1-mx) (x-1)], so.

a) f(x)=-f(-x)=-loga[(1-mx) (x-1)]=loga[(1-mx) (x-1)], m=1 (rounded) roll, m=-1

b) f(x) reputation do = loga[(x+1) (x-1)]=loga[1+2 (x-1)].

Let x1 x2 1, so 1+2 (x1-1)-[1+1 (x2-1)]=2(x2-x1) (x1-1)(x2-1) 0, so.

1+2 (x1-1) 1+1 (x2-1), so.

When 0 a 1, loga[1+2 (x1-1)] loga[1+1 (x2-1)], i.e., f(x1) f(x2), so the function is an increasing function;

When a 1, loga[1+2 (x1-1)] loga[1+1 (x2-1)], i.e., f(x1) f(x2), so the function is a decreasing virtual function;

log: The logarithm of base x is a base.

3log: logarithm of base a with x base - log: log: log: base x log: log: y = 3Can be turned into.

logarithm of base x + logarithm of log base x with a base - logarithm of log base y logth of base a log=3, i.e. (logs with a base logarithm of base x) + 3-log logarithm of base y = logarithm of log base a base x, so logarithm of log base y = (logarithm of base x with a base) + 3-3log high logarithm with base a base x.

Let log be based on a and the logarithm of x = t, then log with a as the base and y as the logarithm = t 2-3t+3, so that the logarithm of log with a as the base y is minimized, even if t 2-3t+3 is the smallest, by the nature of the quadratic function.

Knowing that when t= is the minimum value, then t=>0, the source is closed and therefore satisfied, and x=a to the power.

The log is not easy to play.,Hit it with all hands.,It's so tired.。。。

Solution: 6 x = 8

Take the logarithm on both sides.

xln6=ln8

x=ln8/ln6

In addition, ln8 = ln2 = 3ln2, ln2 = (1 3) ln8ln6 = ln2*3=ln2+ln3

Two-formula simultaneous elimination LN2 has.

ln6-ln3=(1/3)

Both sides of LN8 are divided by LN6

1-ln3/ln6=1/3

ln8/ln6

So ln3 ln6 = 1-(1 3)ln8 ln6 i.e. log[6]3=1-x 3

1. LG50+LG2 LG5 (LG) squared (2) call stove = LG (2 * 25) + LG2 (LG5 + LG2) = LG2 + LG25 + LG2 * LG (5 * 2) = LG2 + 2 LG5 + LG2 = 2 (LG2 + LG5) = 2;

Second, the bent chain accompanies log(2)3 log(4)3) (lg2 lg3) =log(2)3+log(2)3 buried log(2)4)*lg2 lg3=[(3 2)log(2)3]*(lg2 lg3)=(3 2)(lg3 lg2)*(lg2 lg3)=3 2

Because f(x)=(log2 x 2)·(log2 x 4), f(x)=xlog2( )xlog2( ).

xlog2（2）·-xlog2（4）=（-x）·（2x）=2x²

Yes, because 3 log x - so the root number 2 x 8 is f(x) maximum 2·64=128 and minimum 2·2=4 because the function increases monotonically from 2 to 8

Hope you help,(*.)

f(x) = [log2 (x 2)]·log2(x 4)]= (log2 x - 1)· (log2 x - 2) = [(log2 x) -3 2)] 2 - 1 4) and -3 log(1 2) x "-1 2 can be reduced to: 1 2 log2 x 3

1《[(log2 x) -3 2)]《3 2 0《[(log2 x) -3 2)] 2《 9 4 -1 4《[(log2 x) -3 2)]"2 i.e.: f(x) maximum value 2 , minimum value -1 4