Can Exponent Values Be Determined with One Equation and Two Variables?

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Discussion Overview

The discussion revolves around whether exponent values can be determined from a single equation involving two variables, specifically in the context of an equation derived from powers of 3. The scope includes mathematical reasoning and exploration of algebraic manipulation.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses doubt that the values of the exponents can be determined from one equation with two variables.
  • Another participant provides a mathematical derivation showing that from the equation $(3^{3v})(27^w)= 81^{12}$, they can simplify to find that $v + w = 16$.
  • This same participant argues that while there is only one equation for $v$ and $w$, the problem does not require solving for each variable individually.
  • The average of $v$, $w$, and 35 is calculated as 17, based on the derived values.
  • A later post acknowledges a previous post, indicating some overlap in contributions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the exponent values can be definitively determined, as there is a mix of doubt and mathematical reasoning presented.

Contextual Notes

The discussion highlights the limitations of having only one equation for two variables, which raises questions about the ability to uniquely determine the values of those variables.

karush
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ok without any calculation I felt it could not be determined since we have one equation with 2 variables
 
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$27= 3^3$ and $81= 3^4$ so $(3^{3v})(27^w)= 81^{12}$
is $3^{3v}(3^{3w})= (3^4)^12$
$3^{3v+ 3w}= 3^{48}$.

$3v+ 3w= 48$ so $v+ w= 16$.

Yes, we have only one equation for v and w but the question does not ask us to solve for v and w.

The average of v, w, and 35 is $\frac{v+ w+ 35}{3}= \frac{16+ 35}{3}= \frac{51}{3}= 17$.
 
helps to write it out rather than quickly assume things
 
Country Boy said:
$27= 3^3$ and $81= 3^4$ so $(3^{3v})(27^w)= 81^{12}$
is $3^{3v}(3^{3w})= (3^4)^12$
$3^{3v+ 3w}= 3^{48}$.

$3v+ 3w= 48$ so $v+ w= 16$.

Yes, we have only one equation for v and w but the question does not ask us to solve for v and w.

The average of v, w, and 35 is $\frac{v+ w+ 35}{3}= \frac{16+ 35}{3}= \frac{51}{3}= 17$.
thank you
 
sorru this was posted earilier
 

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