#pH=-log_10[H_3O^+]# by definition.

And thus, if #pH=10#, #[H_3O^+]=10^(-10)*mol*L^-1#, (i.e. #pH=-log_10(10^(-10))=-(-10)=10#

And if #pH=8#, #[H_3O^+]=10^(-8)*mol*L^-1#, which is one hundred times more concentrated than the first instance, as required. In other words, if the #DeltapH=2#, there is a #10^2#, i.e. one hundredfold difference in #[H_3O^+]#.

Do not be intimidated by the #log# function. When we write #log_ab=c#, we ask to what power we raise the base #a# to get #c#. Here, #a^c=b#. And thus #log_(10)10=1, #, #log_(10)100=2, ##log_(10)10^(-1)=-1 #. And #log_(10)1=0#.

I acknowledge that I have hit you with a lot of facts. But back in the day A level students routinely used log tables before the advent of electronic calculators. If you can get your head round the logarithmic function you will get it.