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Question

If log3x+log3y=2+log32 and log3(x+y)=2, then


A

x=1,y=8

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B

x=8,y=1

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C

x=3,y=6

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D

x=9,y=3

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Solution

The correct option is C

x=3,y=6


Explanation for the correct answer:

Step 1: Simplify the equations using log properties:

Given: log3x+log3y=2+log32 and log3(x+y)=2

Consider log3x+log3y=2+log32

log3(xy)=2log33+log32[logab=loga+logb;logaa=1]log3(xy)=log332+log32[alogb=logba]log3(xy)=log39+log32log3(xy)=log318xy=18...1

log3(x+y)=2(x+y)=32[logab=cb=ac]x+y=9...2

Step 2: Solve for the required value:

x=9-y

Put the value of x in (1)

(9-y)y=189y-y2=18y2-9y+18=0y2-6y-3y+18=0y(y-6)-3(y-6)=0(y-6)(y-3)=0y=6,y=3

Consider,

y=6x=9-y=9-6=3y=3x=9-y=9-3=6

According to the given options the correct combination is x=3,y=6

Hence option(C) i.e. x=3,y=6 is correct.


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