SOLUTION: 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
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Question 22964: 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)
Found 2 solutions by rapaljer, venugopalramana:
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I would have to agree with the tutor, since I found two solutions when I solved it myself. I don't entirely understand your notation, but I think you may have lost a solution when you went from a quadratic equation to a linear equation:
(logx 3)^2 - 3logx 3 = 4
logx 3 - 3 = 4 / (logx 3)
It might have been safer if you had set it equal to zero, and factored it.
[ logx 3 - 4] [logx 3 + 1] = 0
This way, you too will get two solutions.
R^2 at SCC
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE MY COMMENTS BELOW
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) OK.I SHALL USE YOUR NOTATION TO SHOW SQUAREROOT
(logx 9) / (logx V""x) - logx 3 = 9 + 6( (logx x) / (logx V""3) )...GOOD.YOU TOOK EVERY LOG TO BASE X
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6( 1 / ( (logx 3) / 2 ) )....VERY GOOD
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6 / ( (logx 3) / 2 )..VERY GOOD
2logx 9 - logx 3 = 9 + 6 ( 2 / logx 3 ) .....VERY GOOD
2logx 9 - logx 3 = 9 + 12 / logx 3 .......VERY GOOD
logx 3 ( 2logx 9 - logx 3 - 9 ) = 12 ......VERY GOOD
logx 3 ( 4logx 3 - logx 3 - 9 ) = 12
logx 3 ( 3logx 3 - 9 ) = 12 EXCELLENT
3(logx 3)^2 - 9logx 3 = 12 ....CORRECT
(logx 3)^2 - 3logx 3 = 4.........CORRECT...YOU HAVE DONE EXCEPTIONALLY WELL UP TO THIS .NOW BEST WAY O PROCEED IS..... FOR YOUR EASE OF UNDERSTANDING PUT
LOGX 3 = Y..THEN WE HAVE...Y^2-3Y-4=0=Y^2-4Y+Y-4=0
Y(Y-4)+1(Y-4)=0
(Y-4)(Y+1)=0..
Y=4 OR =-1
LOGX 3=4..........................OR............... LOGX 3=-1
X^4=3........................OR.........X^-1=3
X=3^(1/4).....................OR..........X=3^-1=1/3
NOW LET US SEE THE MISTAKE YOU DID
logx 3 - 3 = 4 / (logx 3)...OK
logx 3 - 3 = 1/4 logx 3.....HOW?THIS IS THE MISTAKE...OK...
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)
I AM VERY HAPPY AT YOUR SINCERE ATTEMPT AND CORRECT ANALYSISALMOST TILL THE END.YOU WILL SURELY GO UP WITH THIS APPROACH.NOW LET ME GIVE YOU A TIP TO DO THIS EASIER..WHILE USING THE FORMULA
TAKE LOGX Y=LOG Y/LOG X DROPPING THE BASE WHICH COULD BE ANY COMMON STANDARD BASE..SO THE INITIAL PROBLEM BECOMES
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
LOG 9/LOG SQRTX-LOG 3/LOG X=9+6*(LOG X/LOG SQRT3)..AND PROCEED ..THIS WILL SIMPLIFY THE WOKING AS YOU ARE NOT TO WORRY ABOUT THE BASE.
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