FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: - x^[log a to the base 2] + a^[log x to the base 2] = 2a^2 I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
Last edited by mr fantastic; May 31st 2010 at 02:22 AM.
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Originally Posted by arijit2005 FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: - x^[log a to the base 2] + a^[log x to the base 2] = 2a^2 I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it???? Dear arijit2005, Although you may not be familiar there's a logarithm identity, This will help you to solve the problem.
Last edited by Sudharaka; May 31st 2010 at 05:38 PM.
Completely wrong here, see above and below posts. It might be prudent for this post to be deleted since it contributes nothing
Last edited by e^(i*pi); May 31st 2010 at 02:24 AM.
Originally Posted by e^(i*pi) By the change of base rule: Should be easy enough to find x in terms of a from there. I have used base e but any base will work Dear e^(i*pi), There's a mistake in your first step;
Originally Posted by arijit2005 FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: - x^[log a to the base 2] + a^[log x to the base 2] = 2a^2 I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it???? Originally Posted by Sudharaka Dear e^(i*pi), There's a mistake in your first step; Of course you're right, how on Earth did I not spot that >.<
Hello arijit2005 Originally Posted by arijit2005 FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: - x^[log a to the base 2] + a^[log x to the base 2] = 2a^2 I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it???? We have to solve for : If in doubt, get rid of logs as soon as you can - they're nasty things! So let . Then . So the equation becomes: Grandad
Hi everyone, Another approch: If you use the logarithmic identity that I had mentioned in my previous post; Hope this will help you.
Last edited by Sudharaka; May 31st 2010 at 05:37 PM.
Originally Posted by Grandad Hello arijit2005We have to solve for : If in doubt, get rid of logs as soon as you can - they're nasty things! So let . Then . So the equation becomes: Grandad WOW.. Thanks.. That's a nice way.. Why the hell didn't I think of that??? I need more practice.
Originally Posted by Sudharaka Hi everyone, Another approch: If you use the logarithmic identity that I had mentioned in my previous post; Hope this will help you. Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula By the way, is it or ??? Will base "a" change?? You actually wrote Are you sure the becomes ???
Hello arijit2005 Originally Posted by arijit2005 Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula By the way, is it or ??? Will base "a" change?? You actually wrote Are you sure the becomes ??? Yes, it should be You can prove this using the same substitution that I used (so that you can get rid of those nasty logs!): Let and Then and So Grandad
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