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1/(x log abc to the base a^x)+1/(y log abc to the base b^y)+1/(z log abc to the base c^z) a)0 b)abc c)xyz d)1

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(a/log abc to the base a^bc)+(b/log abc to the base b^ca)+(c/log abc to the base c^ab) = a)-abc b)abc c)1/abc d)1

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Hope it helps.

bc/log abc to the base a^a + ca/log abc to the base b^b + ab/log abc to the base c^c= a)abc b)1 c)-abc d)1/abc

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Thank you.

(1/(1+log bc to the base a))+(1/(1+log ac to the base b))+(1/(1+log ab to the base c))

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(3+log 343 to the base 10) /(2+1/2 log 49/4 to the base 10 +1/3 log (1/125) to the base 10) = a)6 b)3 c)12 d)9

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ln(ln x)=0 then x= a)0 b)1 c)e d)1/e

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If log N base a = x then the correct option is a)a^N =x b)x^N =a c)a^x =N d)x^a =N

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If a^x =N then the correct option is a)log a base N =x b)log a base x =N c)log a base N =x d)log N base a =x

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Log (1/49) to the base 7= A)1 B)0 C)-1 D)2

If 2 log x base 3 – log (x+6)=1 then x =

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