Math problems with logarithmic exponents in the first year of high school

Simplification: (log4 3 + log8 3) (log3 2 + log9 2) - log1 2 4 root number 32

log4 3 + log8 3) (log3 2 + log9 2) - log1 2 4 root number 32

log2^6+log2^6)(log3^2+log3^4)-log2^（-4）4√2

2*6log2)(6log3)-log2 (-4)2 (5 2)12(log2)(6log3)-log2 (-3 2)12(log2)(6log3)+3 2log2y=log1 2 (2x-1) domain under the root number.

y=√log1/2^(2x-1)

log1/2^(2x-1) ≥0

log1/2^(2x-1) ≥log1

1/2^(2x-1) ≥1

2^(2x-1) ≤1=2^0

2x-1≤0

x 1 2 Good luck.

Jilin Wang Qing llx

log(a n) (b m)=(m n)loga b,loga b+loga c=loga (bc),loga b*logb a=1 basic common aa. Simplify(log(2 2) 3+log(2 3) 3)(log3 2+log(3 3) 2)-log(2 -1) (2*32) 2)=(5log2 3)(4log3 2)+6log2 2=26. It's a basic question when it's annoying to type on a mobile phone, and it's not going to be a liberal arts study.

Square the known and expand it to the 2y power of x + 2 + x to the -2y power.

So the answer is 2 times the root number 2-4

log2[2^(3/2)]≈log2(

log9(25)=log3(5) log3[3^(3/2)]≈log3(

log9(25)< 3/2 =68.Let 2 x=t

then y=t 2-3t+3=(t-3 2) 2+3 4 x [1,7].

2≤t≤128

then its value range is [1,16003].

-a + 1/(a+b)

If you want to know the specific process, you can ask me.

Note: The second part is the X power of multiplying A up and down!

Multiply a x up and down at the same time, and then simplify it to get the result.

1+a＾-x）/2（1-a^-x）

1+1/a＾x）/2（1-1/a^x）

1/a^x(a^x+1))/2(1/a^x(a^x-1)（1+a^x）/2(a^x-1)

In fact, it is to convert a -x into 1 a x and then divide and simplify.

Let y=2 x, the original equation becomes y-1 y=2, i.e., y-2y-1=0, and we get.

y=2^x=(2±√8)/2=1±√2

But when y=1-2<0, the formula is obviously meaningless.

So y=2 x=1+ 2

i.e. x = 2 (1 + 2).

If you are satisfied, welcome to adopt, thank you.

If you have any questions, please feel free to ask.