-6.1.12 solve for x 3^{2x}=105

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Discussion Overview

The discussion revolves around solving the equation \(3^{2x} = 105\) for the variable \(x\). Participants explore different methods and representations of the solution, including logarithmic transformations and base changes.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes rewriting the equation using natural logarithms, arriving at \(x = \frac{\ln{105}}{\ln{9}}\).
  • Another participant suggests a different approach, stating \(x = \frac{\ln{105}}{2\ln{3}}\), noting that both methods yield the same result due to the relationship between logarithms.
  • A third participant introduces the idea of expressing \(x\) as \(\frac{1}{2}\log_3{(105)}\), implying a preference for using base 3 logarithms.
  • Some participants question the necessity of changing the logarithm base, arguing that the base of the exponential equation should dictate the logarithm base used.
  • Concerns are raised about calculator limitations, with one participant noting the absence of a base 3 logarithm key on their calculator.
  • Another participant emphasizes the desire for an exact answer rather than a numerical approximation, questioning the use of calculators in this context.

Areas of Agreement / Disagreement

Participants express differing opinions on the best approach to solving the equation, with no consensus on the preferred method or the necessity of changing the logarithm base. The discussion remains unresolved regarding the optimal strategy for finding the solution.

Contextual Notes

Some participants highlight the limitations of calculators in handling specific logarithm bases, which may affect their approach to the problem. Additionally, there is an emphasis on the preference for exact answers over decimal approximations.

karush
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solve for $\ x \quad 3^{2x}=105$
rewrite $x\ln3^2=x\ln{9}=\ln{105}$
hence $x=\dfrac{\ln{105}}{\ln{9}}$

can this be reduced more except for decimal
 
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I would have done this a little differently:
2x ln(3)= ln(105)

x= ln(105)/2ln(3).

Of course, since 9= 3^2, ln(9)= 2ln(3) so that is the same as your answer. Whether it is "simpler" or not is a matter of personal choice.
 
Why not just $\displaystyle x = \frac{1}{2}\log_3{\left( 105 \right) }$?
 
change of base then..
 
karush said:
change of base then..

I'm asking, why are you even bothering to change the base at all? The base of your exponential equation is 3, surely the best base for your logarithm would therefore also be 3...
 
But my calculator doesn't HAVE a "logarithm base 3" key!
 
Country Boy said:
But my calculator doesn't HAVE a "logarithm base 3" key!

And why are you bothering to use a calculator? Surely they would want an exact answer...
 

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