At 7:30 it is known that log x = log 5/log 2. Fine. So x =10^(log 5/log2) = (10^(log 5)}^(1/log 2), and because 10^(log 5) = 5 ex definitio, so x = 5^(1/log 2). The author blunders along various unneeded conversions for 5 minutes to arrive at the same result.
Since 8=2^3 and 5!=120, 8^log(x)-2^log(x)=120 is equivalent to (2^log(x))^3-2^log(x)-120=0. This is a cubic equation in 2^log(x), with roots 2^log(x)=5 or 2^log(x)=(-5-sqrt(71)i)/2 or 2^log(x)=(-5+sqrt(71)i)/2. Assuming that we are interested in real solutions, it follows that x=5^(1/log(2)). The exact value depends on the base of the logarithm. The author of the clip doesn't specify the base, so we assume natural logarithm, and so x is approximately 10.1953128981289.
Wasn't specified at beginning that only real x is sought. So, not rejecting complex solutions @6:25, we get, w/ "e" Euler's number, "ln" natural log & "i" imaginary unit, m=(-5±i√71)/2=2√6·e^(±iθ), where cosθ=-5/(4√6) & sinθ=√71/(4√6) ⇒ θ≅.67π≅120.7° Then, the other 2 roots of x are x=10^(log₂m)=m^(1/log2)=(2√6)^(1/log2)·e^(±iθ/log2). Isn't that good-looking?
it depends on whether log(2) means log base 10 or natural log. In higher level math classes, log always means natural log, but in High School and probably some lower level college classes, it could be base 10.
At 7:30 it is known that log x = log 5/log 2. Fine. So x =10^(log 5/log2) = (10^(log 5)}^(1/log 2), and because 10^(log 5) = 5 ex definitio, so x = 5^(1/log 2). The author blunders along various unneeded conversions for 5 minutes to arrive at the same result.
Since 8=2^3 and 5!=120, 8^log(x)-2^log(x)=120 is equivalent to (2^log(x))^3-2^log(x)-120=0. This is a cubic equation in 2^log(x), with roots 2^log(x)=5 or 2^log(x)=(-5-sqrt(71)i)/2 or 2^log(x)=(-5+sqrt(71)i)/2. Assuming that we are interested in real solutions, it follows that x=5^(1/log(2)). The exact value depends on the base of the logarithm. The author of the clip doesn't specify the base, so we assume natural logarithm, and so x is approximately 10.1953128981289.
Wasn't specified at beginning that only real x is sought. So, not rejecting complex solutions @6:25, we get, w/ "e" Euler's number, "ln" natural log & "i" imaginary unit,
m=(-5±i√71)/2=2√6·e^(±iθ), where cosθ=-5/(4√6) & sinθ=√71/(4√6) ⇒ θ≅.67π≅120.7°
Then, the other 2 roots of x are
x=10^(log₂m)=m^(1/log2)=(2√6)^(1/log2)·e^(±iθ/log2).
Isn't that good-looking?
let u=2^(logx) , u^3-u-120=0 , (u-5)(u^2+5u+24)=0 , u=5 , / complex , u^2+5u+24 / , 2^(logx)=5 , logx*log2=log5 ,
logx=log5/log2 , x=10^(log5/log2) , test , 8^(log5/log2)-2^(log5/log2)=120 , OK , x=~ 209.859 ,
(8)^2log^(x)^2 ➖ (2)^2log(x)^2 64logx^2 ➖ 4^log^x^2 = 60log^{x^0+x^0 ➖ }=60log^x^1=60^log^x^1 5^12logx^1 5^3^4log^x^1 5^1^3^4^log^x^1 1^1^3^2^2log^x^1 1^3^1^2log^x^1 3^2log^x (logx ➖ 3logx+2).
I’m with you until 5^1/log2. But that comes out to 10.195 when I do it. What am I doing wrong?
it depends on whether log(2) means log base 10 or natural log. In higher level math classes, log always means natural log, but in High School and probably some lower level college classes, it could be base 10.
@@johnlv12 thanks. That makes sense.