Oxford University Admission Exam Question

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@hafeezrehman8692
@hafeezrehman8692 13 сағат бұрын
Nice 👍
@superacademy247
@superacademy247 12 сағат бұрын
Thanks for the appreciation! 😊💕I'm glad you enjoyed it! 💯😁
@EC4U2C_Studioz
@EC4U2C_Studioz 13 сағат бұрын
I don't think you needed to use the power rule for logs as it’s implied from canceling the base with the log of the same base leaving with whatever is in the exponent on one side of the equation.
@farancient
@farancient 11 сағат бұрын
x-y = 2, x = 2 + y 2^(2+y) - 2 ^y = 4 2^y * (2^2 - 1) = 4 2^y = 4/3 y = (log 4-log3) / log2 = 0.4150 X = 2.4150
@davidshen5916
@davidshen5916 12 сағат бұрын
4=2^X-2^Y=2^Y(2^(X-Y)-1)=2^Y(2^2-1)=3*2^Y,2^Y=4/3, Y=LOG(4/3)/LOG(2)=2-LOG(3)/LOG(2),X=4-LOG(3)/LOG (2)
@nikolayplatnov5148
@nikolayplatnov5148 9 сағат бұрын
Oxford admission to what ? To the "course of equity and verieties " 😂😂😂
@ChavoMysterio
@ChavoMysterio 57 минут бұрын
2^x-2^y=4 x-y=2 -----> y=x-2 2^x-2^(x-2)=4 2^x-¼(2^x)=4 4(2^x)-2^x=16 3(2^x)=16 2^x=⅓(2⁴) 2^(x-4)=⅓ x-4=log_2(3^-1) x-4=-log_2(3) x=4-log_2(3) ❤ y=4-log_2(3)-2 y=2-log_2(3) ❤
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