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

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The discussion centers on solving the equation \(3^{2x} = 105\). Participants derive the solution using logarithmic properties, arriving at \(x = \frac{\ln{105}}{\ln{9}}\) and \(x = \frac{\ln{105}}{2\ln{3}}\). They debate the simplicity of these forms and the appropriateness of changing the logarithm base, ultimately suggesting that using base 3 is more intuitive given the original equation. The conversation emphasizes the preference for exact answers over decimal approximations.

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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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