add. maths!!!~~~thzxs

1
{[log(a+3)] / (log9)} - (logb / log9) = c+0.5
[log(a+3)] - (logb) = log9*(c+0.5)...(1)
[log(a-3) / log3] + (logb / log 3) = c -1
[log(a-3) ] + (logb) = log3*(c -1)...(2)
(1)+(2)
[log(a+3) ]+[log(a-3) ]= log9*(c+0.5)+ log3*(c -1)
log[(a+3)(a-3) ]= 2log3*(c+0.5)+ log3*(c -1)
log[a^2-9 ]= log3*[2(c+0.5)+(c -1)]
log[a^2-9 ]= log3*[3c]
log[a^2-9 ]= log3*[log(27^c)]/log3
log[a^2-9 ]= [log(27^c)]
a^2 = 9 + 27^c
(2) Prove that y = x^2 + 2kx + 2(k^2) + k +1 is always positive for any real values of k.
Discriminant
=(2k)^2-4(2(k^2) + k +1)
=4k^2-8k^2-4k-1
=-(4k^2+4k+1)
=-(2k+1)^2
<=0
so
y = x^2 + 2kx + 2(k^2) + k +1 is always positive for any real values of k.