Question 22964
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HELP =) !

resolving the Question 2570 (p200 - 250) : log of 9 to the base square root of x - log of 3 to
the base x = 9 + 6 log of x to the base square root of 3 a tutor found x = 3^(1/4) or x = 1/3
I only found one answer x = 3^(1/4) 

Who is right ? And where is the (my?) flaw ?
Thanks a lot !!!

Here is my answer :

logV""x(9) - logx(3) = 9 + 6logV""3(x)                       
(logx 9) / (logx V""x) - logx 3 = 9 + 6( (logx x) / (logx V""3) )       
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6( 1 / ( (logx 3) / 2 ) )   
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6 / ( (logx 3) / 2 ) 
2logx 9 - logx 3 = 9 + 6 ( 2 / logx 3 )       
2logx 9 - logx 3 = 9 + 12 / logx 3             
logx 3 ( 2logx 9 - logx 3 - 9 ) = 12  
logx 3 ( 4logx 3 - logx 3 - 9 ) = 12  
logx 3 ( 3logx 3 - 9 ) = 12   
3(logx 3)^2 - 9logx 3 = 12     
(logx 3)^2 - 3logx 3 = 4
logx 3 - 3 = 4 / (logx 3)
logx 3 - 3 = 1/4 logx 3
logx 3 - 3 = logx 3^(1/4)
logx 3 - logx 3^(1/4) = 3
logx (3/3^(1/4) ) = 3
x^3 = 3/3^(1/4)                              
x^3 = 3^(3/4)                  
x^3 = 3^( (1/4) * 3 )
x^3 = ( 3^(1/4) )^3
x = 3^(1/4) 

Is this: {{{log (sqrt(x), (9)) - log (x, (3)) = 9 + 6*log (sqrt(3), (x))}}}?

Just like the other tutor, I find it difficult to follow your solution. So, it's difficult to point out your flaw(s).

Your tutor is correct. There're 2 solutions: {{{highlight(x = matrix(2,1, " ", 3^(1/4)))}}} and {{{highlight(x = 3^(- 1) = 1/3)}}}
Is your line 11 {{{(log (3, (x)))^2 - 3*log (3, (x)) = 4}}}   OR   {{{(log (x, (3)))^2 - 3*log (x, (3)) = 4}}}?
The correct equation should be: {{{4(log (3, (x)))^2 + 3*log (3, (x)) - 1 = 0}}}. This'll lead to the 2 CORRECT solutions.</pre>