In fact, you can use two different changes of variables v=(2/5)^x as you did or u=(5/2)^x as I did. And both give same equation v+1/v=20 or u+1/u=20. So, the results shall be opposite as x1=-x2. And that’s the case, v1 = 10+3sqrt(11) and v2=10-3sqrt(11)=1/v1. So, you can say after computing x1 that x2=-x1. Easy.
An easy way to verify that is squaring both numbers and comparing them. (10)² = 100 (3sqrt(11))² = 9*11 = 99 100 > 99, therefore, 10 > 3sqrt(11). Then, 10-3sqrt(11) is positive.
At 6:30, the ecuation is : v² - 20v +1 = 0. Still, the values for a, b and c are correct.
V2-19v 7.35 minutes equations not satisfied -20v+v will e subtract . In this equation c=0
Good❤
Error in solution at 6.17 min1/v×v = 1 not v.
Yes! its mistake 😊
Equation v²+v=20v at 6:15 is incorrect. Supposedly it is v²+1=20v
In fact, you can use two different changes of variables v=(2/5)^x as you did or u=(5/2)^x as I did. And both give same equation v+1/v=20 or u+1/u=20.
So, the results shall be opposite as x1=-x2. And that’s the case, v1 = 10+3sqrt(11) and v2=10-3sqrt(11)=1/v1. So, you can say after computing x1 that x2=-x1. Easy.
There is a mistake. The answer same, but negative.
Where? I chucked step by step its correct throughout.
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A^n=1/a^-n (5/2)^x=1/(5/2)^-x
In denominator power should be negative.
There is no check if 10-sqrt(11) is positive. Not obvious with casual inspection (needs calculator, old fashioned log table not precise enough)
An easy way to verify that is squaring both numbers and comparing them.
(10)² = 100
(3sqrt(11))² = 9*11 = 99
100 > 99, therefore, 10 > 3sqrt(11). Then, 10-3sqrt(11) is positive.
Ah yes of course. It's been over 40 years since I last dabbled in mental gymnastics and had forgotten all these tricks 😢
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It's Ans is 1
?
Because I take log both side
You joking?
5:54 1/V x V = V ?