Three logarithmic function learning questions: 200 points are added

Detailed explanation of the answer:1First, calculate a=4

How do you calculate a=4? To use the following formula, logam+logan=logamn (to explain, a is the base m, n is the true number, and mn is the product.)

According to the known conditions and this formula.

f(m)+f(n)=f(mn).

From the * formula and f( 3-1) + f ( 3 + 1) = know a = 4 by * formula, m = 17-1, n = 17 + 1, a = 4 to get f( 17-1) + f ( 17 + 1) = 2

2.To do this kind of question, you should take into account the problems of "the denominator is not zero" and "the true number is greater than zero".

Therefore, the formula in parentheses is greater than zero and not one, and the range of the solution is (1,2)(2,3)3If the question is problematic and you want to find a defined domain instead, you should do this:

Doing such problems takes into account such problems as "true numbers are greater than zero" and "monotonicity of logarithmic functions".

It is known from the meaning of the title. The thing in parentheses is "greater than 1 (according to the logarithmic function monotonicity, and of course don't forget to consider" the true number is greater than zero"Oh! The exponent here refers to "something in parentheses")

It is derived. log2x is less than 1 (2 is the base, x is the true number) and the domain is defined as (0,2) (according to the monotonicity of the logarithmic function, "true x is greater than zero").

The first is 2 ... There's an F missing in front.。。。 According to the formula, you can go to the plus sign in the middle of the left formula.

At the same time, multiply the root number 3 plus or minus 1. Left = loga [(root number 3-1) (root number 3+1)]. It becomes loga2=1 2...

I know that a is 4. Then find log4 16 which is 2

The second bracket is greater than zero and not one... It counts out as (1,2)u(2,3).

I didn't understand the third question. How about two value ranges... If you want to define the field, it is (0,2).

1. Because loga(x 2+4)+loga(y 2+1)=loga5+loga(2xy-1).

So (x +4) (y +1) = 5 (2xy 1).

The result is x y +4 + x +4 = 10xy-5

It can be transformed into x y 6xy+9+4y +x 4xy=0, i.e., (xy 3) + 2y x) = 0

So 2y-x=0, i.e. y x=

log8y/x=﹣1/3

2. When x belongs to (0,1), f(x)=3 x-1

And because f(x) is an odd function, when x (-1,0), -x (0,1), then f(x) = f(-x) = 3 (-x)+1

Because -4 = log1 3 81 log1 3 36 log1 3 27 = 3, log1 3 36 (-4, -3).

Because f(x+3 2) = f(x-3 2), i.e. f(x) = f(x+3).

Let x0=log1 3 36

Since x0 (-4,-3), then x0+3 (-1,0), so f(x0)=f(x0+3)= 3 (-x0)+1

So f(log1 3 36) = 35

3. Let t=logax x, then logx ax=1 t

logax x+logx (ax)^2=logax x+2logx ax=t+2/t＞0

So t=logax x 0

logax x =(lnx)/(lnax)=(lnx)/(lnx+lna)＞0

When lnx 0, i.e., x 1, lnx + lna 0, so x 1 a

When lnx 0, i.e. 0 x 1, lnx + lna 0, so 0 x 1 a

In summary, if a 1, then x 1 or 0 x 1 a

If 0 a 1, then x 1 a or 0 x 1

— If you don't understand, you can ask again or hi