{log3(x)+2^(y+2)=1
{log3(x^2)+2^y=3

Solve the system

Question

{log3(x)+2^(y+2)=1
{log3(x^2)+2^y=3

Solve the system

alekos · Answer

well I suppose you can start by subtracting as the equations stand

welshfella · Answer

yes that would eliminate 2^y

welshfella · Answer

are those logs to the base 3?

czarluc · Answer

yes

alekos · Answer

then solve for x after some log rule manipulation

czarluc · Answer

so the equation would be -log3(x) +2^2=-2?

alekos · Answer

I'll give you the first step and then I want you to try the second

czarluc · Answer

okay

czarluc · Answer

the term 2^(y+2) is a correction

czarluc · Answer

the "+2" is part of the exponent of 2

alekos · Answer

stand by

alekos · Answer

OK bottom equation becomes

2logx + 2^y = 3

and now we multiply the top equation by 2 and then subtract the equations to eliminate 2logx

czarluc · Answer

then the equation becomes 4^(y+2)-2^y=-1

alekos · Answer

No that's way out, but I just tried to solve for y and I get the log of a -ve number which is undefined, so that can only mean you've given me the wrong equations.
I reckon the 1 and 3 need to be swapped

czarluc · Answer

Its the exact equations given, maybe it's just undefined

alekos · Answer

If that's the case then there is no solution.  Have they given you an answer?

czarluc · Answer

No its only an example. the system is just inconsistent

alekos · Answer

Yeah, that's another way of putting it