There is a satisfies 2a 2 a 5, b satisfies 2b lg x 10 lg2 5, and a b is found

I'll give it a try. I've seen the original question, and LZ still made a mistake.

The original question is 2a+2 a=5

2b+2lg(b-1) lg2=5, find a+b and let me prove the original problem.

Solution: 2a+2 a=5 (a-1)+2 (a-1)=3 22b+2lg(b-1) lg2=2b+2log2 (b-1)=5 (b-1)+log2(b-1)=3 2

Convert the yuan, let a=a-1, b=b-1

2^a=3/2-a

log2 b=3/2-b

So let m(a,2 a),n(b,log2 b)y1=2 x, y2=log2 x, y3=3 2-xy1 y3=a,y2 y3=b

Since y1,y2 are inverse functions of each other, y1,y2 are symmetric with respect to y=x.

Take the intersection of y3 and y=x p(3 4,3 4).

The slope relation is known, y3 y=x.Therefore pm=pn

Therefore p is the midpoint of mn, a+b=2p=3 2

a+b=a+b+2=7/2

b satisfies 2b+lg(x-10) lg2=5??How can there be an x.

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A satisfies 2A+2 A=5, B satisfies 2B+LG(X-10) LG2=5, and A+B7 satisfies 2B+LG(B-10) LG2 = . 2b+2lg(b-1) lg2=5, find a+b

2LG(B-A 2)=LGA+LGB Then: A>0,B>0,B>A 2

lg(b-a/2)²=lgab

b-a/2)²=ab

b²-ab+a²/4=ab

b²-2ab+a²/4=0

4b²-8ab+a²=0

a²-8ab+4b²=0

Same as dividing b a b) -8a b + 4 = 0

Solution: a b = 4 2 3

Because b>a 2, a b<2

So: a b = 4-2 3

Have fun! I hope it can help you, if you don't understand, please hi me, I wish you progress in your studies! o(∩_o

2lg(b-a/2)=lg(b-a/2)^2lga+lgb=lgab

So (b-a 2) 2=ab

b^2+a^2/4-ab=ab

b^2+a^2/4=2ab

Both sides of the equation are divided by b squared.

The square of 1 + (a 2b) = 2a b

Let a b=x

then 1+x 2 4=2x

x^2-8x+4=0

The solution gives x=(8+4*root3) 2=4+2*root3 or x= 4-2*root3

lga+lgb=lgab

2lg(b-a2)=lg(b-a2) 2 Note b-a2>0(b-a2) 2=ab

b^2-ab+a^2/4=ab

Divide by b 2 on both sides

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

a b=4+2*3 (1 2)[rounded].

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

A -2ab-9b = 0 and both sides are divided by b (b ≠0) to get (a b) -2(a b) - 9 = 0

Solution: a b = 1 10

A and B have the same name. So a b 0

a/b=1+√10

LG(A +AB-6B)-LG(A +4AB+15B)=LG[(A +AB-6B) (A +4AB+15B)]=LG (up and down divided by B at the same time).

lg=lg(1+2√10+10+1+√10-6)/(1+2√10+10+4+4√10+15)

lg[(6+3√10)/(6√10+30)=lg[(2+√10)/(2√10+10)=lg(√10/10)

lg10(-1/2)

It's good to make use of the squared difference formula, why subtract? (a+b) 2-(a-b) 2=[a+b+a-b][a+b-a+b]=2a*2b=4ab=-2, the squared difference formula will be more efficient than subtraction.

LGA+LGB=LGAB on the left

Right 2lg(a-2b)=lg(a-2b)2 left Right, then ab=(a-2b)2

Divide by b2 on both sides

Simplified as: a b = (a which hall b-2)2

another a b = x

i.e. x=(x-2)2

x2-5x+4=0;

x-1)(x-4)=0

x=1 or x=4

When x=1, Li Qi, Fu Pao a-2b=-b<0; Not eligible.

Therefore a b = 4

The landlord's condition "4a 2+ab 2=4ab" is incorrect, it should be "4a 2+9b 2=4ab".

From 4a 2+9b 2=4ab(a>0), we can see that adding 12ab to both sides of b>0 yields: (2a+3b) 2=16ab【(2a+3b) 4] 2=ab

Take LG on both sides of the above equation to obtain: LG(2A+3B) 4=(LGA+LGB) 2

I hope it helps you and I wish you progress in your studies!

Q: I pushed back.

Find out if the landlord's question is wrong.

It should be 4a +9b = 4ab

Then there is. 4a²+12ab+9b²=16ab

2a+3b）²=16ab

2a+3b)/4】²=ab

Go logarithm on both sides.

lg【（2a+3b）/4】²=lgab

2lg（2a+3b）/4=lga+lgb

LG(2A+3B) 4=(LGA+LGB) 2 proof completed.

I hope the landlord will check the topic.

First check whether the question is wrong, 4a 2 + ab 2 = 4ab, isn't it possible to directly make an appointment with an a???