+0  
 
0
84
7
avatar+23 

Im struggling lol

 Dec 6, 2020
 #1
avatar+23 
+1

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

 Dec 6, 2020
 #3
avatar+112000 
+2

I do have an answer for  question1 

 

 let A=3^x   and   B=2^x

 

Do the substitution and then solve as a quadratic.

 Dec 6, 2020
 #4
avatar+31506 
+4

Here's the answer to the first one

 Dec 6, 2020
 #7
avatar+23 
+1

i was curiouse how you simplified 2^2x+1 to 2(2/3)^x

raphae1  Dec 6, 2020
 #5
avatar+114098 
+3

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

 

 

cool cool cool

 Dec 6, 2020
 #6
avatar+114098 
+3

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 x=  1

 

√(9log 3 x)  -  1  -  log 9 x=  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

 

 

cool cool cool

 Dec 6, 2020

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