How to solve for x when it's in exponent?

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Discussion Overview

The discussion revolves around methods for solving equations where the variable x appears in the exponent. Participants explore different interpretations of an example equation and the use of logarithms in the solution process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the clarity of the original equation, suggesting that the formatting may lead to different interpretations of what is in the exponent.
  • It is proposed that taking logarithms of both sides is a necessary step to isolate x, with the base of the logarithm being flexible.
  • One participant humorously notes the difficulty of using a cash machine to solve the problem, implying that the equation's complexity may be overwhelming.
  • A specific example is provided to illustrate the process of solving for x using logarithms, demonstrating the steps involved in isolating x.
  • There is a suggestion that the original poster may not have fully understood the mathematical concepts involved, leading to a humorous take on the situation.

Areas of Agreement / Disagreement

Participants express differing views on the clarity and correctness of the original equation, with no consensus on its interpretation. While there is agreement on the method of using logarithms, the specific application and understanding of the equation remain contested.

Contextual Notes

Limitations include potential misunderstandings of the equation's structure and the assumptions made about the variable x's placement within the exponent. The discussion does not resolve these ambiguities.

Who May Find This Useful

This discussion may be of interest to individuals seeking to understand the application of logarithms in solving exponential equations, as well as those curious about the interpretation of mathematical expressions.

sarah2529
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I'm just wondering because I found this crazy picture somewhere on the net.

http://photos-h.ak.fbcdn.net/hphotos-ak-snc3/hs071.snc3/13845_1261921983558_1094945438_30798960_833983_n.jpg
 
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Log based 264
 
I see no "x in the exponent" here. Or is that "264^15/99xyz" supposed to be "264^(15/99xyz)" rather than "(264^15)/(99xyz)"?

What about the "*(2n+1)- (pi+ phi)"? Is that in the exponent or not? Looks like the person who posted that on the internet didn't really know what he was doing.

If everything is in the exponent then, like fatra2 said, take logarithms of both sides. If not, subtract the part that is not in the exponent from both sides and then take the logarithm.

You don't NEED to take logarithms base 264 (although that gives the simplest result). Logarithms to any base will do.
 
It probably isn't at all difficult to solve for x, with the exponent interpreted in any way
I'd say your problem is finding the log base 264 button on that cash machine.
 
Haha I would need more than 30 secs just to press all the buttons needed to solve for x, let alone figuring out the answer and what the question actually is - from what it says, I would interpret it as

[tex]y=\frac{264^{15}.2xyzn}{99}+1-(\pi+\phi)[/tex]

But it could obviously mean many other forms. And like hallsofivy has said, the person is obviously clueless about this kind of stuff and was leaning much towards the humour of this picture rather than the accuracy of its question.
 
To answer the basic question, "How to solve for x when it's in the exponent?":

You would need to isolate the x-term on one side of the equation, take the logarithm of both sides of the equation (any base, as long as they're the same), move x from exponent of a logarithm to coefficient of a logarithm [itex](\log{3^X} = X \log 3)[/tex]<br /> <br /> Example:<br /> <br /> [itex]6561 = 3^X[/tex]<br /> [itex]\ln{6561} = \ln{3^X}[/tex]<br /> [itex]\ln{6561} = X\ln3[/tex]<br /> [itex]X = \frac{\ln6561}{\ln3}[/tex]<br /> [itex]X = 8[/tex]<br /> [itex]\therefore 6561 = 3^8[/tex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]
 

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