Even if you are only looking for real number solutions, the argument for case 2 as presented is insufficient, because the equation at that stage does not contain any logarithms. You need a stronger auxiliary property to reject case 2, such as that a^x=b with a>0, b
2 errors. In the video, the left-hand side of the equation has minus sign between the 2 terms, whereas in the thumbnail they're added. 2ndly, since wasn't specified that x is real, case 2 has a complex solution. 7^(-1/x)=-4 ⇒ x=-ln7/ln(-4)=-ln7/ln(4e^(iπ))=-ln7/(2ln2+iπ)=ln7·(-2ln2+iπ)/[4(ln2)²+π²] where we've chosen branch cut of logarithm such that arguments of complex numbers lie in the interval (-π,π], so that the log is single-valued & consistent w/ the real solution.
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At this point, the problem is quite simple. Thanks 4 all teacher.
You're welcome 💕🥰🤩✅
Even if you are only looking for real number solutions, the argument for case 2 as presented is insufficient, because the equation at that stage does not contain any logarithms. You need a stronger auxiliary property to reject case 2, such as that a^x=b with a>0, b
For negative values of x: (log(base a))(-x)=log(base a)(x)+(pi*i/ln(a)).
Logs of negative numbers are defined. Just use the complex plane.
in case anyone is interested, the numerical solution is -1.209
X=log7/log(1/5) , test , 1/49^(log(1/5)/log7)- 1/7^(log(1/5)/log7)=25-5 , --> 20 , OK ,
2 errors. In the video, the left-hand side of the equation has minus sign between the 2 terms, whereas in the thumbnail they're added. 2ndly, since wasn't specified that x is real, case 2 has a complex solution.
7^(-1/x)=-4 ⇒ x=-ln7/ln(-4)=-ln7/ln(4e^(iπ))=-ln7/(2ln2+iπ)=ln7·(-2ln2+iπ)/[4(ln2)²+π²]
where we've chosen branch cut of logarithm such that arguments of complex numbers lie in the interval (-π,π], so that the log is single-valued & consistent w/ the real solution.
It was a mistake now rectified. Thanks for letting me know 🙏🙏🙏