Wiedząc że log2≈ 0,3010, log3≈ 0,4771, log6≈0,7781 i log7≈ 0,8451, oblicz: c) 2log21 - log28 d) log 3√2/7 e)log √7√2 f) log 1/4√3 g) 2[log6-log3-log2+log7]

c)  2log21 - log 28 = log(21² : 28) = log168 = log(3 × 7 ×8 ) = log3 +log7 + log8 = log3 + log7 + log2³ + 3log2  = 0,4771 + 0,8451 + 3×0,3010 =  0,4771 + 0,8451 + 0,9030 = 2,2252 d)  log3√2/7 = log(3×2^{1/2} :7^{-1}) = log3 + log2^{1/2} - log7^{-1} = log3 + 1/2log2 - (-log7) = log3 + 1/2log2 + log7 = 0,4771 + 1/2*0,3010 + 0,8451 = 0,4771 + 0,1505 + 0,8451 = 1,4727 e)   log√[7√2] = log√7 + log√[√2] = log7^{1/2} + log2^{1/4}= 1/2log7 + 1/4log2 = 1/2×0,8451 + 1/4×0,3010 = 0,42255 + 0,07525 = 0,4978 f)  2[log6-log3-log2+log7] = 2(0,7781 - 0,4771-0,3010+0,8451) = 2×0,8451 = 1,6902