Indices. solving for the pronumerals: 3^²x−¹/2^x+y = 9/4, a^²x−¹b^y+² =√b/a

[cookie101]

Indices. solving for the pronumerals: 3^²x−¹/2^x+y = 9/4, a^²x−¹b^y+² =√b/a

can someone help me with these questions im really stuck and its due this week.

solve the following for the pronumerals.

3^²x−¹/2^x+y = 9/4

a^²x−¹b^y+² =√b/a

thank you!

 

[Deleted member 4993]

can someone help me with these questions im really stuck and its due this week.

solve the following for the pronumerals.

3^²x−¹/2^x+y = 9/4

a^²x−¹b^y+² =√b/a

thank you!

Looks like your problem is (if not - please correct your post):

\(\displaystyle \displaystyle{\dfrac{3^2x^{-1}}{2^{x+y}} \ = \dfrac{9}{4}}\)

\(\displaystyle \displaystyle{a^2x^{-1}b^{y+2} \ = \dfrac{\sqrt{b}}{a}}\)

It should have been written as:

3^² * x−¹/2^(x+y) = 9/4

a^² * x−¹ * b^(y+²) = √b/a


Follow the method shown in your previous post: http://www.freemathhelp.com/forum/threads/95970-HELP-solving-for-one-unknown-in-the-power

If you are still stuck - come back and show us exactly how far you have proceeded and where you are stuck.