What is the solution to the logarithmic equation with different bases?

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Discussion Overview

The discussion centers around solving the logarithmic equation $\log_9{(x+1)}+3\log_3{x}=14$. Participants explore various approaches to manipulate the equation, particularly focusing on changing the base of the logarithms involved.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant suggests changing the base of $\log_9{(x+1)}$ to $\dfrac{1}{2}\log_3{(x+1)}$ or $3\log_3{x}$ to $6\log_9{x}$, indicating that either transformation leads to a complex polynomial equation.
  • Another participant mentions a numerical approximation for $x$ as approximately 80.8579, expressing surprise at the base change method.
  • Several participants reiterate the change of base formula, $\log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$, providing a derivation of this formula.
  • One participant questions the origin of the factor $\dfrac{1}{2}$ in the transformation, prompting clarification from others.
  • It is noted that the factor $\dfrac{1}{2}$ arises because 3 is the square root of 9, which is explained by a participant.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the change of base formula and its application, but there is no consensus on the best approach to solve the equation or the implications of the resulting polynomial.

Contextual Notes

The discussion involves assumptions about logarithmic properties and transformations, with some participants expressing uncertainty about the implications of their manipulations.

karush
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$\tiny{KAM}$
$\log_9{(x+1)}+3\log_3{x}=14$
ok not sure as to best approach to this
assume change the base 9?
 
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change

$\log_9(x+1)$ to $ \dfrac{1}{2}\log_3(x+1)$

or …

$3\log_3{x}$ to $6\log_9{x}$

either way will result in a nasty 7th degree polynomial equation that will require technology to solve
 
by w|A $x≈80.8579$

but I didn't know you could change the base like that
 
change of base formula …

$\log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$
 
skeeter said:
change of base formula …

$\log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$

derivation …

let $x = \log_b{a} \implies a = b^x$

$\log_c{a} = \log_c{b^x}$

$\log_c{a} = x \cdot \log_c{b}$

$\dfrac{\log_c{a}}{\log_c{b}} = x = \log_b{a}$
 
skeeter said:
change

$\log_9(x+1)$ to $ \dfrac{1}{2}\log_3(x+1)$

or …

$3\log_3{x}$ to $6\log_9{x}$

either way will result in a nasty 7th degree polynomial equation that will require technology to solve
where did the $\dfrac{1}{2}$ come from
 
$\log_9(x+1) = \dfrac{\log_3(x+1)}{\log_3{9}} = \dfrac{\log_3(x+1)}{2} = \dfrac{1}{2} \log_3(x+1)$
 
skeeter said:
$\log_9(x+1) = \dfrac{\log_3(x+1)}{\log_3{9}} = \dfrac{\log_3(x+1)}{2} = \dfrac{1}{2} \log_3(x+1)$
well that's good to know
 
The "1/2" is due to the fact that 3 is the square root (1/2 power) of 9.
 

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