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3^2y + 6(3^y)=27
because some bases are not the same (6(3^y))
we'll use another means.
let 3^y=P
anywhere you see 3^y, replace it with P
note:3^2y=(3^y)²
P²+6P=27
P²+6P -27=0
solve quadratically.
factors of -27P² that will give me +6P
9×3=27 , 9-3=6
.:+9p-3p==>(if you multiply it, it'll give you -27P², Also if you add it it'll give you+6p)
this part is critical.
Let's equate +9p-3p for+6p in p²+6P -27=0
p²+9p-3p -27=0
group into 2 with bracket.
(p²+9p)(-3p -27)=0
bring out common factor
p(p+9)-3(p+9)
group the numbers outside together and bring one p+9 in another bracket
(p-3)or(p+9)
p-3=0. p+9=0
p=3. p=-9
so p is either+3 or -9
but we'll go with positive integers+3
p=3
Remember, 3^y=p
now p=3, substituting it you have
3^y=3
in indices 3¹=3, 3⁰=1, 3²=9 etc
so 3^y=3¹
bases(3) cancel out
y=1
E
The question is more of indices, logarithm is not stated in the question. The answer is supposed to be 3
