Question 829796
Log(base 10)16-log(base10)(2t)=log(base 3)10
{{{log((16))-log((2t))=log(3,(10))}}}
{{{log((16/2t))}}}=2.0959
10^(2.0959)≈16/2t
124.7096≈16/2t
2t≈0.1283
t≈0.0641