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

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The discussion centers on determining exponent values using a single equation with two variables. It establishes that from the equation $(3^{3v})(27^w)= 81^{12}$, one can derive that $v + w = 16$. The conversation emphasizes that while there is only one equation, it does not require solving for individual values of v and w. Additionally, it calculates the average of v, w, and 35, resulting in 17. The conclusion reinforces that the problem can be approached without needing to find specific values for v and w.
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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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