Find x if
log1218 = log24x + 1. log24x+1 + 4
A) 0
B) 1
C) 2
D) None of these
Answers
Answered by
0
Let I=log18log12⋅log54log24+5(log18log12−log54log24)I=log18log12⋅log54log24+5(log18log12−log54log24). Also, let log3=xlog3=x and log2=ylog2=y.
Then,
I=log32⋅2log22⋅3.log33⋅2log23⋅3+5(log32⋅2log22⋅3−log33⋅2log23⋅3)=2x+y2y+x⋅3x+y3y+x+5(2x+y2y+x−3x+y3y+x)I=log32⋅2log22⋅3.log33⋅2log23⋅3+5(log32⋅2log22⋅3−log33⋅2log23⋅3)=2x+y2y+x⋅3x+y3y+x+5(2x+y2y+x−3x+y3y+x)
=6x2+5xy+y2+10x2+35xy+15y2−15x2−35xy−10y2(2y+x)(3y+x)=x2+5xy+6y2x2+5xy+6y2=1
Then,
I=log32⋅2log22⋅3.log33⋅2log23⋅3+5(log32⋅2log22⋅3−log33⋅2log23⋅3)=2x+y2y+x⋅3x+y3y+x+5(2x+y2y+x−3x+y3y+x)I=log32⋅2log22⋅3.log33⋅2log23⋅3+5(log32⋅2log22⋅3−log33⋅2log23⋅3)=2x+y2y+x⋅3x+y3y+x+5(2x+y2y+x−3x+y3y+x)
=6x2+5xy+y2+10x2+35xy+15y2−15x2−35xy−10y2(2y+x)(3y+x)=x2+5xy+6y2x2+5xy+6y2=1
Answered by
0
hey mate
1is the answer
☺✔✔✔✔✔✔
1is the answer
☺✔✔✔✔✔✔
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