Log(9-t)+log(9+t)=0.3

Question

Can someone pls hell me

Answer 1

log(9-t)+log(9+t)=0.3,

log[(9-t)(9+t)]=0.3,

log(81-t²)=0.3.

We need to know the log base to solve this. Let's assume the base is 10.

81-t²=10^0.3=1.9953 approx.

t²=81-1.9953=79.0047, t=±√79.0047=±8.8885 approx. t=8.8885 or -8.8885.

(So 9-t=0.1115 or 17.8885 and 9+t=17.8885 or 0.1115.)