stack_equiv_find_step(ex, ex1) example Working line by line, fully simplify {@first(ta)@} showing all your working.

[[input:ans1]] [[validation:ans1]]

]]> 1 0.1 0 2018120500 ta:[lg(25,5),stackeq(lg(5^2,5)),stackeq(2*lg(5,5)),stackeq(2*1),stackeq(2)] [[feedback:prt1]] {@ta@} 0 0 1 Correct answer, well done.]]> Your answer is partially correct.]]> Incorrect answer.]]> . cross 1 i cos-1 [ ans1 equiv ta 15 1 0 firstline 0 1 0 0 1 2 prt1 1.0000000 0 w1:stack_equiv_find_step(lg(5^2,5), ans1) 0 EquivFirst ans1 ta 0 = 0.5000000 1 prt1-1-T = 0.0000000 -1 prt1-1-F 1 EqualComAss last(ans1) last(ta) 0 + 0.5000000 2 prt1-2-T - 0.0000000 -1 prt1-2-F You have not quite finished simplifying your answer! Please carry on working.

]]> 2 AlgEquiv emptyp(w1) true 0 + 0.0000000 -1 prt1-3-T Please put in an intermediate step showing how you deal with the logarithm of powers of a number.

]]> - 0.0000000 -1 prt1-3-F 1 ans1 ta prt1 1.0000000 0.0000000 prt1-3-F 2 ans1 [lg(25,5),stackeq(lg(5^2,5)),stackeq(2*lg(5,5)),stackeq(2*1)] prt1 0.5000000 0.1000000 prt1-2-F 3 ans1 [lg(25,5),stackeq(2)] prt1 1.0000000 0.0000000 prt1-3-T