please answer me indices
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reduce everything in powers of 3 and 2.
9^y=3^(2y)
27y=3^(3y)
our qn becomes:
{3^(2y)*3^2*3^y*3^3y}/{3^(3x).2^3}=1/27
add the powers of 3.
3^(6y+2-3x)=1/27=3^(-3)
equate both the powers and get the value:-)
9^y=3^(2y)
27y=3^(3y)
our qn becomes:
{3^(2y)*3^2*3^y*3^3y}/{3^(3x).2^3}=1/27
add the powers of 3.
3^(6y+2-3x)=1/27=3^(-3)
equate both the powers and get the value:-)
rerek:
thanks bro. but i can't understand after 3^6y+2-3x=3^-3
that is where is 2^3 ??
please answer me bro
when i was answering no one was watching,so i felt demotivated to give complete solution
ok please send me answer
i will edit the answer.
ok bro thank you
im not able to solve it.sorry.
its ok i am happy for your effort to solve my question amd to answer my queries.thank you so much and so kind of you bro.
no problem bro/sis.I will try to solve other questions of you if possible:)
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