Elitmus
Exam
Numerical Ability
Log and Antilog
if z=(log(base a)(log(base b)(a)))/(log(base b)(log(base a)(b)))
Read Solution (Total 8)
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- Answer should be -> z= -ln(base a) b
- 9 years agoHelpfull: Yes(16) No(3)
- z=(log(base a)(log(base b)(a)))/(log(base b)(log(base a)(b)))
=> z=[log(base b)(a))/log(a)]*[log(b)/(log(base a)(b))]
=> z=[log(a)/{log(a)*log(b)}]*[{log(b)*log(a)}/log(b)]
=> Z=log(a)/Log(b)
=> Z=log(base b) (a) .....Ans - 9 years agoHelpfull: Yes(6) No(3)
- plz upload the full problem.
- 9 years agoHelpfull: Yes(3) No(1)
- Hold your breath :p
z=(log(base a)(log(base b)(a)))/(log(base b)(log(base a)(b)))
Let log b base a = x
then, log a base b = 1/x
=> log 1/x base a / log x base b
=> -log x base a / log x base b
=> - 1 / log a base x / 1 / log b base x
=> - log b base x / log a base x
Ans = - log b base a [ using property ]
Cheers :D - 7 years agoHelpfull: Yes(3) No(0)
- heheeee mujhe nai mallooom I guess1
- 9 years agoHelpfull: Yes(0) No(0)
- z=(b^3/2)/(a^3/2)
- 9 years agoHelpfull: Yes(0) No(0)
- answer would be = -logb^a or log a^b
- 7 years agoHelpfull: Yes(0) No(0)
- please submit genuine solution
- 7 years agoHelpfull: Yes(0) No(0)
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