If `3^x = 4^(x-1)` then x=(A) `(2og_32)/(2log_32-1)` (B) `2/(2-log_23)` (C) `1/(1-log_43)` (D) `(2log_23)/(2log_23-1)`

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If 3^(x)=4^(x-1), then x=(2log_(3)2)/(2log_(3)2-1)(b)(2)/(2-log_(2)3)(1)/(1-log_(4)3)(d)(2log_(2)3)/(2log_(2)3-1)

log_(2)(x+1)-log_(2)(3x-1)=2

(1)/(log_(2)x)+(1)/(log_(3)x)+......(1)/(log_(43)x)=(1)/(log_(43)!x)

(log_(8)17)/(log_(9)23)-(log_(2sqrt(2))17)/(log_(3)23)=

I 1,log_9(3^(1-x)+2), log_3(4.3^x-1) are in A.P. then x equals (A) log_34 (B) 1-log_43 (C) 1-log_34 (D) log_4 3

(1)/(log_(2)(n))+(1)/(log_(3)(n))+(1)/(log_(4)(n))+....+(1)/(log_(43)(n))

(1)/("log"_(2)n) + (1)/("log"_(3)n) + (1)/("log"_(4)n) + … + (1)/("log"_(43)n)=

log_(2)(x^(2)-1)=log_((1)/(2))(x-1)

(log_(8)17)/(log_(9)23) - (log_(2sqrt(2))17)/(log_(3)23) is equal to