What is the value of 2 ^ log4 (2 - √ 3) + 3 ^ log9 (2 + √ 3)`

What is the value of 2 ^ log4 (2 - √ 3) + 3 ^ log9 (2 + √ 3)`

2^log4(2-√3) + 3^log9(2+√3)
=2^log2^2(2-√3) + 3^log3^3(2+√3)
=2^(1/2[log2(2-√3)]) + 3^(1/2[log9(2+√3)])
=2^1/2*(2-√3)+3^1/2*(2+√3)
=2√2-√6+2√3+3

Given that log95 = a, log37 = B, try to use a and B to represent the value of log359 (represented by a and b)

log9 (5)=a,
log3 7=1/2*log9(7)=b
log9(7)=2b
log35 9
=1/log9(35)
=1/[log9(5)+log9(7)]
=1/(a+2b)

Log9] 5 = m, log3] 7 = n
Yes, please give me the process

log35】9=1/log9]35
log9]35=log9]7+log9]5
log9]7=2log3]7=2n
So log35] 9 = 1 / (2n + m)

5 times of log9 = a, 7 times of log9 = B, 9 times of log35,

35 times of log9 = 5 times of log9 + 7 times of log9 = a + B
9 times of log35 = 1 / 35 times of log9 = 1 / (a + b)

Given log95 = a, log97 = B, find the value of log359
Online, etc

From log95 = a, log97 = b
Log9 35 = log9 5 + log9 7 = a + B
So log359 = 1 / log935 = 1 / (a + b)

If log3 (4) = a, log3 (5) = B, then log9 (16 / 25)=

log9(16/25)=log3(4/5)=log3(4)-log3(5)=a-b

|log3(5)-2|+log9(25)+(1/32)^1/5

|log3(5)-2|+log9(25)+(1/32)^1/5
=2-log3(5)+log3(5)+1/2
=5/2

The order of three numbers a = 3 ^ 0.7, B = 0.7 ^ 3, C = log3 (0.7) is
Log3 (0.7) is the base of log and 7 is the true number

A = 3 ^ 0.7 ＞ 3 ^ 0 = 1 [because y = 3 ^ x is an increasing function]
0 < B = 0.7 ^ 3 < 0.7 ^ 0 = 1 [because y = 0.7 ^ x is a decreasing function]
C = log3 (0.7) < log3 (1) = 0 [because y = log3 (x) is an increasing function]
Therefore: a > b > C
I don't know how to ask~

If a = log3 10, B = log3 7, then 3 (a)_ The value of (2b) Note, (a)_ 2b) is 3_

3^(a-2b)=(3^a)/(3^b)^2=10/(7^2)=10/49

Log3 ^ 7 = a, log7 ^ 4 = B. use a and B to represent log14 ^ 84

Log3 ^ 7 = a, log7 ^ 4 = B, log7 ^ 2 = B / 2log14 ^ 84 = log14 ^ 7 + log14 ^ 12 = 1 / (log7 ^ 14) + 2log14 ^ 2 + log14 ^ 3 = 1 / (log7 ^ 7 + log7 ^ 2) + 2 / log2 ^ 14 + 1 / log3 ^ 14 = 1 / (1 + B / 2) + 2 / (log2 ^ 7 + 1) + 1 / (log3 ^ 7 + log3 ^ 2) = 2 / (2 + b) + 2 / (2 / B + 1) + 1 / (a + AB / 2) = 2 / (2 + b)