Solving simultaneously..
x = 3^y - - - - - - eqn1x=3y−−−−−−eqn1
x = 1/2(3 + 9y) - - - - - - eqn2x=12(3+9y)−−−−−−eqn2
Look at the common value in both equations..
xx is the common, hence we equate both together..
Having..
3^y = 1/2(3 + 9y)3y=12(3+9y)
3^y = (3 + 9y)/23y=3+9y2
Cross multiplying..
3^y/1 = (3 + 9y)/23y1=3+9y2
2xx 3^y = 3 + 9y2×3y=3+9y
6^y = 3 + 9y6y=3+9y
Log both sides..
log6^y = log(3 + 9y)log6y=log(3+9y)
Recall the law of logarithm -> log6^y = x, ylog6 = x→log6y=x,ylog6=x
Therefore...
ylog6 = log(3 + 9y)ylog6=log(3+9y)
Divide both sides by log6log6
(ylog6)/(log6) = log(3 + 9y)/(log6)ylog6log6=log(3+9y)log6
(ycancel(log6))/cancel(log6) = log(3 + 9y)/(log6)
y = (log(3 + 9y))/log(6)
y = (cancel (log)(3 + 9y))/(cancel (log)(6))
y = (3 + 9y)/6
Cross multiplying..
y/1 = (3 + 9y)/6
6 xx y = 3 + 9y
6y = 3 + 9y
Collect like terms
6y - 9y = 3
-3y = 3
Divide both sides by -3
(-3y)/(-3) = 3/-3
(cancel(-3)y)/cancel(-3) = 3/-3
y = -3/3
y = - 1
Substitute the value of y into eqn1 to get x
x = 3^y - - - - - - eqn1
x = 3^-1
Recall in indices, x^-1 = 1/x
:. x = 1/3
Hence the values are rArr x = 1/3, y = -1
Hope this helps!
