can anyone solve this ?
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using proptery of log
x=loga/log2a
y=log2a/log3a
z=log3a/4a
it gives xyz = loga/log4a = log(a) of base 4a
and 1 can be written as log4a base 4a
since log (x) base x =1
xyz+1 = loga base 4a + log4a base 4a
= log(4a.a) base 4a
since logx+logy of same base =logxy
and u can write it in many forms like
xyz+1 = 2[log2a base 4a] too
x=loga/log2a
y=log2a/log3a
z=log3a/4a
it gives xyz = loga/log4a = log(a) of base 4a
and 1 can be written as log4a base 4a
since log (x) base x =1
xyz+1 = loga base 4a + log4a base 4a
= log(4a.a) base 4a
since logx+logy of same base =logxy
and u can write it in many forms like
xyz+1 = 2[log2a base 4a] too
Divini:
what is the value for log 2a to the base 4a
options are..a) 2yx b)2xy c)2zx d)none of these
sorry a)2yz
u should have mentioned it earlier that we have to ans. in form of xyz
wait let me type the ans.
sorry
see the last ans.! it came 2(log2a base 4a) means log2a/log4a !! and log 2a =log3a.y(2nd eqn) and write log 4a as log3a/z( from 3rd eqn) ...log3a cancels out and u get 2yz as answer
thank you
if it helped!! u can rate it! (only if u get that)#happy to help
My pleasure!
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