Really Hard Log Problems

btw please show how you did it, i want to learn how to do this type of log because im kinda interested

I do have an answer for  question1 

 let A=3^x   and   B=2^x

Do the substitution and then solve as a quadratic.

Here's the answer to the first one

5.  We can write

(1/3)  ( log x / log 2)   + (log x /log 2)^(1/3)   =  4/3

Let Logx /log 2   =  a   

And we have

(1/3)a +  a^(1/3)   = 4/3      rearrange as

a^(1/3)    =  4/3  -  (1/3)a

a^(1/3)  =  (4 - a)   / 3        cube both sides

a  =   (4 - a )^3   /  27    

Note that this is true when  a =  1

So  a =    log x / log2   = 1      →    x  = 2

2   .√ ( log ₃ x⁹)  -  4log ₉ √(3x)  = 1

√ (log ₃ x⁹)  -  log ₉  √(3x)⁴    =1

√(log ₃x ⁹) -   log ₉ (3^2  x²)   =  1

√ (log ₃ x⁹) - log ₉ 3^2   -   log ₉ x²   =  1

√(9log ₃ x)  -  log ₉ 9  -  log ₉ x² =  1

√(9log ₃ x)  -  1  -  log ₉ x² =  1

√(9log ₃ x)    - 2 log ₉ x =  2

3 ( log 3  / log x)^(1/2)  -  2 (log x / log 9 )  =  2

3 ( log 3 / log x)^(1/2)   - 2  (log x  / log 3^2)   =2

3( log 3 / log x)^(1/2)   -  2  (log x /(2 log 3) )    = 2

3 ( log3/ log x)^(1/2)  - log x /log 3     = 2

3  (log 3/logx)^(1/2)    -  1 / (log 3/logx)    = 2

Let  (log 3/log x)^(1/2)   =  a

Then   log3/logx   =  a^2

So we have

3a   - 1/a^2   = 2        multiply through by a^2

3a^3   -1  =2a^2

3a^3  - 2a^2   = 1       and it's  clear that  this will be true when a  = 1

So      (log 3 / log x)^(1/2)    =1          square both sides

log 3 / log x  =  1

x  = 3