we know that 2^y is non-zero real number, so we let 2^x - 2^y = 2^y(2^x/2^y - 1) since the second equaiton shows that x-y=2; we have 2^y(2^2 - 1) = 3*2^y; as the first equation shows, it's equal to 4, we have 3*2^y=4 i.e. 2^y=4/3; we take log base 2 on both side, log2^y=log4/3 i.e. ylog2=log4-log3 i.e. y=2-log3, and as second equation shows, x=2+y, we have x=2+(2-log3)=4-log3.
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.
Nice 👍
Thanks for the appreciation! 😊💕I'm glad you enjoyed it! 💯😁
we know that 2^y is non-zero real number, so we let 2^x - 2^y = 2^y(2^x/2^y - 1) since the second equaiton shows that x-y=2; we have 2^y(2^2 - 1) = 3*2^y; as the first equation shows, it's equal to 4, we have 3*2^y=4 i.e. 2^y=4/3; we take log base 2 on both side, log2^y=log4/3 i.e. ylog2=log4-log3 i.e. y=2-log3, and as second equation shows, x=2+y, we have x=2+(2-log3)=4-log3.
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.
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
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)
Oxford admission to what ? To the "course of equity and verieties " 😂😂😂
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) ❤