I paused the vid at the start and tried to mentally solve it, I realized the second terms was just the negative of the first term so it simplified into the first term = 1, from then on its easy to get x^2 - 25 = 0
... Dear friend Newton, In very short my solution strategy: Let LOG(b3)(X^2 - 16) = T, and in general LOG(b 1/A)(B) = LOG(b A)(1/B), a very cool rule don't you think, and very easy to prove! LOG(b 1/3)(1/(X^2 - 16)) = LOG(b 3)(X^2 - 16) = T ... so LOG(b 2)(T) - LOG(b 1/2)(T) = 2 ... LOG(b 2)(T) + LOG(b 2)(T) = 2 ... 2LOG(b 2)(T) = 2 ... LOG(b 2)(T) = 1 ... T = 2 (nice result don't you think too?), so next: LOG(b 3)(X^2 - 16) = 2 ... X^2 - 16 = 3^2 = 9 ... X^2 = 25 ... X1 = - 5 v X2 = 5 ... and finally after checking these candidate solutions in the original equation, we can conclude that they are both valid, so S = { - 5, 5 }. We could also only check one solution, because every imput is being squared, so it doesn't matter which solution we use, 5 or - 5! Friend Newton, final remark, the blackboard tricked us again by letting out twice parentheses right at the beginning of your announcement, and on the black board, take a good 2nd look (lol) ... it remains for me to wish you a very very pleasant weekend my friend ... Only the best, Jan-W
That's unfair! I thought the bars of log2 are covering the entire left side and ended up with an exponential equation with a linear component, there should be another bar covering log3 inside of log2
![Logarithm Change of Base Formula & Solving Log Equations - Part 1 - [7]](http://i.ytimg.com/vi/BIgf3pH9PTY/mqdefault.jpg)
I paused the vid at the start and tried to mentally solve it, I realized the second terms was just the negative of the first term so it simplified into the first term = 1, from then on its easy to get x^2 - 25 = 0
Excellent one bro
No. I just ... I just can't, not today. My head hurts just looking at that.
🤣🤣🤣
At 6:47: 2 (log B / log 2) = 2 log B / log 2 = 1 => B = 2
Haha 😄. Smart
... Dear friend Newton, In very short my solution strategy: Let LOG(b3)(X^2 - 16) = T, and in general LOG(b 1/A)(B) = LOG(b A)(1/B), a very cool rule don't you think, and very easy to prove! LOG(b 1/3)(1/(X^2 - 16)) = LOG(b 3)(X^2 - 16) = T ... so LOG(b 2)(T) - LOG(b 1/2)(T) = 2 ... LOG(b 2)(T) + LOG(b 2)(T) = 2 ... 2LOG(b 2)(T) = 2 ... LOG(b 2)(T) = 1 ... T = 2 (nice result don't you think too?), so next: LOG(b 3)(X^2 - 16) = 2 ... X^2 - 16 = 3^2 = 9 ... X^2 = 25 ... X1 = - 5 v X2 = 5 ... and finally after checking these candidate solutions in the original equation, we can conclude that they are both valid, so S = { - 5, 5 }. We could also only check one solution, because every imput is being squared, so it doesn't matter which solution we use, 5 or - 5! Friend Newton, final remark, the blackboard tricked us again by letting out twice parentheses right at the beginning of your announcement, and on the black board, take a good 2nd look (lol) ... it remains for me to wish you a very very pleasant weekend my friend ... Only the best, Jan-W
Wow! Your keenness is top-notch. I had to look closely to find it 😂😂😂😂.
You are correct. The blackboard is a trickster. I just noticed the missing parenthesis. Enjoy your weekend!
... We're in this together Newton, so we are correct! (lol) ... take care and happy Sunday, Jan-W@@PrimeNewtons
Awesome video, I appreciate how clean your chalk board is for every video!
nice explanation!
First time I was actually able to solve it myself :)
Me too
Good job!
👏🏻
For firest person tanks for an other video master...
Wow black guy doing math
I would have just combined the logs given that logb^c(a) = 1/clogb(a).
That's unfair! I thought the bars of log2 are covering the entire left side and ended up with an exponential equation with a linear component, there should be another bar covering log3 inside of log2
You are correct. I need to fix that ASAP. My apologies.
Вау, это же номер из ЕГЭ по математике