Solve the simultaneous equations #log_a(x+y) = 0# and #2log_ax = log(4y +1)# for #x# and #y# ?

Question

2137 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-15T14:55:50

Solution is #(-2+sqrt7,3-sqrt7)#

As #log_a(x+y)=0#, #x+y=1# as log of #1# is always zero. (A)

Further #2log_ax=log_a(4y+1)=>log_ax^2=log_a(4y+1)# (B)

i.e. #x^2=4y+1#

as from A, #x=1-y#, putting this in B, we get

#(1-y)^2=4y+1#

or #1-2y+y^2=4y+1#

or #y^2-6y+2=0#

and using quadratic formula

#y=(6+-sqrt(36-8))/2=3+-sqrt7#

If #y=3+sqrt7#, #x=-2-sqrt7#, but we cannot have #x<0#

and if #y=3-sqrt7#, #x=-2+sqrt7#

Hence solution is #(-2+sqrt7,3-sqrt7)#