Find the solution to the logarithmic function problem and help solve a logarithmic function problem

This equation is simplified using the range of y [0,1] and the properties of logarithms (logarithm is 0 when true number is 1, logarithm is 1 when true number is equal to base number).

When y is 0, (2*x 2+bx+a) (x 2+1)=1 When y is 1, (2*x 2+bx+a) (x 2+1)=3 because, 0<=y<=1, and the logarithm is a monotonic increasing function.

So, 1<=(2*x 2+bx+a) (x 2+1)<=3

The range of the function is [0,1], then.

1<=(2*x 2+bx+a) (x 2+1)<=3 finishing, get: 1<=2+(bx+a-2) (x 2+1)<=31<=(bx+a-2) (x 2+1)<=1x 2+1>0, then.

x^2+1)<=bx+a-2<=x^2+10<=x^2+1-bx-a+2=(x-b/2)^2-(b^2)/4+3-a

0<=x^2+1+bx+a-2=(x+b/2)^2-(b^2)/4-1+a

For the above formula to hold, then.

b^2)/4+3-a=0 (1)

b^2)/4-1+a=0 (2)

The solution is obtained by equation (1) and (2).

a=2b=2,-2

Take the logarithms of xlog3=ylog4=zlog6z=ylog4 and shensanlog6 on both sides

x=ylog4/log3

1 Z-1 x=log6 ylog4-log3 ylog4=log2 ylog4=1 2y

Let x be the year, (84%) x=y, the second question, y=, the solution is x=lg2 (

f(x)=log(1 a)(2-x) is incremented over the defined domain.

0< 1/a <1 a>1

Let 1-x 2=t.

a>1

When t decreases, g(x) decreases.

The monotonic reduction interval is [0,+

If the logarithm of 3 base 4 x the logarithm of base 4 x the logarithm of base 8 m = the logarithm of base 4 base 2, find m

log3 4 xlog4 8 x log8 m =log4 2=1/2

lg4/lg3 lg8/lg4x lgm/lg8=1/2

lgm/lg3=1/2

m = 3 under the quadratic root

Find the logarithm of 4 base 3 x the logarithm of 9 base 2 + the logarithm of 32 base 4 root sign with 2 base.

The logarithm of 32 under the fourth root number at the base of 2 = 5 4

Logarithm of 4 base 3 x logarithm of base 9 2 = lg3 lg4 xlg2 lg9=lg3 2lg2 xlg2 2lg3=1 4

So the result is 3 2

The logarithm of 4 with 2 base x the logarithm of 4 with 4 base x the logarithm of 3 with 4 = 2

The value of a base 2 logarithm + a base 1 2 is a log (a>0 and a≠1) is 0

If a is the logarithm of base 2 = m, and the logarithm of a base 3 = n, find 2m + n power = 12 of a

10 to the 2nd power of 4-lg4 to the power of 5 is equal to 125

It's all about using the bottom change formula, are you satisfied? I've worked so hard, hurry up, don't forget to add points!

log3 4 xlog4 8 x log8 m =log4 2=1/2

lg4/lg3 lg8/lg4x lgm/lg8=1/2lgm/lg3=1/2

m = 3 under the quadratic root

Find the logarithm of 4 base 3 x the logarithm of 9 base 2 + the logarithm of 32 base 4 root sign with 2 base.

The logarithm of 32 under the fourth root number at the base of 2 = 5 4

Logarithm of 4 base 3 x logarithm of base 9 2 = lg3 lg4 xlg2 lg9=lg3 2lg2 xlg2 2lg3=1 4

So the result is 3 2

The logarithm of 4 with 2 x the logarithm of 3 with 4 x the logarithm of 4 with 3 = 2 with a as the logarithm of 2 + 1 with a as the base 1 2 as the logarithm (a>0 and a≠1) is 0, if the logarithm of a is base 2 = m, the logarithm of a base 3 = n, find 2m + n power = 12 of a

10 of 2-LG4 to the 5th power, etc. Agree.

There is a major difficulty manual on it, and the formula for changing the bottom is telling you that the answer is equal to harming you, and understanding it yourself is the most important thing. Remembering to solve logarithmic functions, that is, remembering those five formulas, practicing repeatedly, is almost the same.

1.Solution: The function f(x) is a composite function, which is composed of two functions, f(x)=log, the logarithm of u(x) with a base, and the x-power-1 of u(x)=a.

Because when a>1, the logarithm of u(x) based on a by f(x)=log is an increasing function, and the x-power-1 of u(x)=a is also an increasing function.

So the logarithm of the function f(x)=log with a base (a's x-power-1) is an increasing function.

When 0, the function f(x)=log is based on a (a's x-power-1) and the logarithm is an increasing function.

Same increase and difference decrease) The logarithm of the function f(x) = log with a as the base (a's x power -1) is the increasing function.

2.Because the square of x is 0, the square of 3-x is 3

And because the square of 3-x is a true number, the square of 0<3-x is 3

When the true number belongs to the interval (0,3), the domain of the logarithm of the function y=log with a base of 2 (the square of 3-x) is (negative infinity, log is a logarithm with a base of 2 and a base of 3].

3.When x belongs to (negative infinity, 1), f(x) = (1 2) x squared (axis of symmetry x=0).

So f(x) decreases monotonically on (negative infinity, 0) and increases monotonically on (0,1).

When x belongs to [1,9), f(x)=1+log with 3 as the base of x increases monotonically on x belongs to [1,9].

So f(x) decreases monotonically on (negative infinity, 0) and increases monotonically on (0,1) and [1,9].

2x 2-5x-3>0 (x>3 or x "a1 2), composite function f(y) = loga(y) a>1

y= 2x 2-5x-3, using the same increase and difference decrease, a>1, f(y) is the increase, which is divided into two categories.

When x>3, y(x) increases and increases together, so the original function increases.

x《一1 2时, y(x) subtracts, heterogeneously subtracts, so the original function subtracts.

In this kind of problem, the composite function y=g(f(x)), let u=f(x)=2x-5x-3, then the outer function y=g(u)=log(a)u, the inner function y=f(x), and the increase and decrease of the composite function is determined according to "the same increase and different decrease".

Solution: Find the definition domain: 2x a 5x a 3>0, (2x ten 1) (x a 3) > 0, x "a 1 2 or x > 3, define the city as:

a, a 1 2) u(3, ten), the inner function f(x) = 2x a 5x a 3, 1 opening upward, the axis of symmetry is: x=5 4, f(x) in (a, a 1 2) monotonically decreasing, f(x) in (3, ten) monotonically increasing, a>1, the outer function y=log(a)u is an increasing function, and the decreasing interval of the function obtained by the properties of the composite function is: (a, a 1 2; The increase interval is:

3, x).

Define the domain 2x-5x-3 0

Solve x -1 2, x 3

This parabolic opening is upward.

So the axis of symmetry decreases on the left and increases on the right.

Because a 1 loga(x) is incremented.

So the monotonic interval of the function is the same as the true number.

So the increase interval is (3,+.)

The minus interval is (- 1 2).

solution, f(x) is meaningful.

t=2x2-5 -3>0 i.e. x>3, or x-1 2.

t is increasing at x>3 and f(x) is also increasing.

t decreases at x<-1 2 and f(x) also decreases.

y=log(2x^2-5x-3)

2x^2-5x-3 >0

2x+1)(x-3)>0

x<-1/2 or x>3

Define the domain =(- 1 2) u (3,+

g(x) =2x^2-5x-3

g'(x) = 4x-5

g'(x)= 0

x= 5/4

g''(x) = 4 >0 (min)

Monotony. Increment = [4, +

Decreasing = (- 1 2) u (3,+4].

solution, so that g(x)=2x 2-5x-3=(2x+1)(x-3)g(x)>0, then >3 or x -1 2

g (x)>0,4x-5>0 then >5 4,f(x)=1oga(g(x))a>1 monotonically increasing and.

g(x)-like.

then x>3, f(x), x<-1 2, f(x).

Find the monotonic interval of 2x 2-5x-3 first.

Solution 2x 2-5x-3>0, get x< -1 2 (monotonically decreased), x>3 (monotonically increased).

a>1, so the monotonicity is the same.