Oxford University Admission Interview Tricks.✍️🖋️📘💙

[@prollysine]

log(4)x-log(4)8/log(4)x^2=1 , log(4)x-(3/2)/(2log(4)x)=1 , let u=log(4)x , u^2-u-3/4=0 , u= 3/2 , -1/2 , --> x= 8 , 1/2 , test x=1/2 , log(4)|1/2|-log(1/4)8=-1/2-(-3/2) , -1/2+3/2=1 , OK , test x=8 , log(4)8-log(64)8=-3/2-1/2 , 3/2-1/2=1 , OK ,

[@ManojkantSamal]

^=read as to the power *=read as square root Now explain log x of base 4=logx/log4 =logx/(log2^2)=logx/(2.log2 ) Explain log8 of base x^2=log8/logx^2 =(3log2)/(2logx) Let log2=a, logx=b So, log 8 of base x^2=3a/2b logx of base 4=b/2a As per question (b/2a)-(3a/2b)=1 (b^2-3a2)/2ab=1 b^2-3a^2=2ab b^2+a^2-4a^2=2ab a^2+b^2-2ab=4a^2 (a-b)^2=(2a)^2 So, a-b=2a or b-a=2a=b=3a a-2a=b So, b=-a So, logx=-log2..... Eqn1 logx=(-1)×log2 log x=log{2^(-1)} logx=log (1/2) X=(1/2).........May be Again logx= - log2 X=(-2).......May be If b=3a, then logx=3.log2 logx=log(2^3) So, logx=log8 X=8 Hence numerous answers of 'X' may arises