[SOLVED]-gre.qu.3 exponent values

karush

Well-known member

ok without any calculation I felt it could not be determined since we have one equation with 2 variables

Country Boy

Well-known member
MHB Math Helper
$27= 3^3$ 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 $\frac{v+ w+ 35}{3}= \frac{16+ 35}{3}= \frac{51}{3}= 17$.

karush

Well-known member
helps to write it out rather than quickly assume things

karush

Well-known member
$27= 3^3$ 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

karush

Well-known member
sorru this was posted earilier