so lets deal with 2 first. We can express 2 in terms of log5 by the laws of logs. nlogx=logx^n. re-writing 2 as 2log5=log25 we now have log(x+1)+log5=log25. lets apply a different log law: log(a)-log(b)=log(a/b). Therefore we get log(x+1)=log(25)-log(5)=log(25/5)=log(5). Now we can cancel the logs to get x+1=5 and now solve algebraically giving x=4