pls solve and answer the question in detail If log45 = x and log56 = y, then (A) log46 = xy (B) log64 = xy (C) log32 = 1 2 x y - 1 (D) log23 = 1 2 x y - 1

log45=xlog56=yxy=log45 log56Use logab=logxblogxaxy=loga5loga4 loga6loga5xy=loga6loga4 Use converse of logab=logxblogxaxy=log46log46=xylog22 6=xyUse logam b==1mlogab12log2 6=xylog2 6=2xylog2 2×3=2xyUse logamn=logam+loganlog2 2+log23=2xy1+log23=2xylog23=2xy-1Use of logab=1logba1log32=2xy-1log32=12xy-1

pls solve and answer the question in detail.
If log₄5 = x and log₅6 = y, then
(A) log₄6 = xy (B) log₆4 = xy (C) log₃2 = $\frac{1}{2 x y - 1}$ (D) log₂3 = $\frac{1}{2 x y - 1}$

\log_{4} 5 = x l {og}_{5} 6 = y x y = \log_{4} 5 l {og}_{5} 6 U s e l {og}_{a} b = \frac{l {og}_{x} b}{l {og}_{x} a} x y = \frac{l {og}_{a} 5}{l {og}_{a} 4} \frac{l {og}_{a} 6}{l {og}_{a} 5} x y = \frac{l {og}_{a} 6}{l {og}_{a} 4} U s e c o n v e r s e o f l {og}_{a} b = \frac{l {og}_{x} b}{l {og}_{x} a} x y = l {og}_{4} 6 l o g_{4} 6 = x y l o g_{2^{2}} 6 = x y U s e \log_{a^{m}} b = = \frac{1}{m} \log_{a} b \frac{1}{2} l o g_{2} 6 = x y l o g_{2} 6 = 2 x y l o g_{2} (2 \times 3) = 2 x y U s e \log_{a} (m n) = \log_{a} m + \log_{a} n l o g_{2} 2 + l o g_{2} 3 = 2 x y 1 + l o g_{2} 3 = 2 x y l o g_{2} 3 = 2 x y - 1 U s e o f l {og}_{a} b = \frac{1}{l {og}_{b} a} \frac{1}{l o g_{3} 2} = 2 x y - 1 l o g_{3} 2 = \frac{1}{2 x y - 1}

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ANSWER IS BOTH (A) AND (C)
since through solving both logarithmic function we will get-
let, log₄5=x
      5=4^x or 5=(2²)^x=2^2x    (equation one)
let, log₅6=y
       6=5^y or 6=(2^2x)^y=2^2xy (from equation one)
this can also be written as: 3x2=2^2xy
                                         3=2^2xy/2¹ ;which is equal to 2^2xy-1
                                      so,3=2^2xy-1
NOW CHECK THE OPTIONS
(A) log₄6=xy
      LHS- log_2²6 = 1/2 log₂6 = 1/2 log₂2^2xy=1/2 x 2xy = xy   (since 6=2^2xyand log_aa^b=b x log_aa = b x 1 =b [PROPERTY-1])
      HENCE, LHS=RHS
                                      CORRECT ANSWER
(B) log₆4=xy
     LHS-log₆2²= 2log_2^2xy2 = 2 x 1/2xy= 1/xy   (since 6=2^2xy and 4=2² and log_a^ba =1/b x log_aa = 1/b x 1= 1/b [PROPERTY-2])
      HENCE, LHS IS NOT EQUAL TO RHS
                                       WRONG ANSWER
(C) log₃2=1/2xy-1
      LHS-log₃2= log_2^2xy-12 = 1/2xy-1 x log₂2 = 1/2xy-1 (since in starting it was proved that 3 can be written as 2^2xy-1and PROPERTY-2
is also used)
HENCE , LHS=RHS
                                        CORRECT ANSWER

(D) log₂3 = 1/2xy-1
      LHS- log₂3 =log₂2^2xy-1=2xy-1 x log₂2 = 2xy-1 x 1 = 2xy-1     (since 3 can be written as 2^2xy-1and PROPERTY-1 is also used)
       HENCE, LHS IS NOT EQUAL TO RHS
                                       WRONG ANSWER

pls solve and answer the question in detail. If log45 = x and log56 = y, then (A) log46 = xy (B) log64 = xy (C) log32 = 1 2 x y - 1 (D) log23 = 1 2 x y - 1

pls solve and answer the question in detail.
If log₄5 = x and log₅6 = y, then
(A) log₄6 = xy (B) log₆4 = xy (C) log₃2 = $\frac{1}{2 x y - 1}$ (D) log₂3 = $\frac{1}{2 x y - 1}$