I've been struggling with a question and was determined not to just look at the answer until I knew how to do it. When the Professor got to the using log to solve for x part I did the same for my question and just breathed out a huge sigh of relief when it came back as correct!
In the problem at 33:30 - does this just have no solution because one of the terms in the original has a negative x? If you plug -1/9 into x and then combine the logarithms, you end up with log[(-10/9)/(-1/9)]=1 which becomes log10=1
at 51:44 he says (in my mind, confusingly) "now this is important, divide again by the next number" when you can just do what I feel is clearly stated in the equation, which is divide by (4*log(1.015)) taking care with the parenthesis. I double checked on a calculator and they come out the same
I came back here to remember how to calculate how many days it would take the coronavirus to infect my entire country if it continued to infect people at the same rate lol.
is this the same logarithmic an exponential functions thats in calculus? wanna double-check because it's in my calc 1 syllabus but I cant find it in the calc1 playlist from this channel
At around 26 minutes in when hes explaining that the -3 doesnt work, it appears to me that it does in fact work. When you add the 2 logs together you get -3 x -2 and end up with log base 6 to the first power = 6. That appears to work to me. Please correct me if im wrong
+Crazyhorse2882 He's not saying -3 doesn't output 6, he's saying that the exponent of a logarithm can't be negative. Remember when you convert the logarithm to an exponent the answer to that exponent is the exponent of the logarithm. The only way to get a negative answer to an exponent is if the base is negative, and in logarithms the base is never negative. I'm not the best at explaining, but I'm pretty sure that's why -3 doesn't work.
-3 is a solution to the equation log6 x(x+1) = 1 but it is not a solution to the original equation. When we combined the logs using the product property, we lost the information that x must be >0 which came from log6 x. Whenever we lose information during some transformation, extraneous solutions show up and we have to check.
Jesus christ, you have a gift for teaching. I understand this subject now.
52:00 LOL this guy makes math so fun and so instructive...
I've been struggling with a question and was determined not to just look at the answer until I knew how to do it. When the Professor got to the using log to solve for x part I did the same for my question and just breathed out a huge sigh of relief when it came back as correct!
Excellent explanation. Informative.
In the problem at 33:30 - does this just have no solution because one of the terms in the original has a negative x? If you plug -1/9 into x and then combine the logarithms, you end up with log[(-10/9)/(-1/9)]=1 which becomes log10=1
Best Professor.
at 51:44 he says (in my mind, confusingly) "now this is important, divide again by the next number" when you can just do what I feel is clearly stated in the equation, which is divide by (4*log(1.015)) taking care with the parenthesis. I double checked on a calculator and they come out the same
Nailed it again! Thanks!
Thank you so much! currently taking precalculus and having a hard time. Your videos have helped me so much!
Thank you so much, professor. Your videos help me a lot😊
I came back here to remember how to calculate how many days it would take the coronavirus to infect my entire country if it continued to infect people at the same rate lol.
Thank you soo much, you are amazing.
I hope from you to make on circles when there are 3 points.
awesome........
is this the same logarithmic an exponential functions thats in calculus?
wanna double-check because it's in my calc 1 syllabus but I cant find it in the calc1 playlist from this channel
If someone dislikes this video, they are truly evil in nature.
A brilliant scientific calculator for your students is the : fx-83ES ( CASIO ) Natural Display.
But( -1/9 - 1)/-1/9 is positive?
thank you so much
WE ARE NOT ALLOWED TU USE CALCULATORS IN MY PRECAL CLASS.
было три
At around 26 minutes in when hes explaining that the -3 doesnt work, it appears to me that it does in fact work. When you add the 2 logs together you get -3 x -2 and end up with log base 6 to the first power = 6. That appears to work to me. Please correct me if im wrong
+Crazyhorse2882 He's not saying -3 doesn't output 6, he's saying that the exponent of a logarithm can't be negative. Remember when you convert the logarithm to an exponent the answer to that exponent is the exponent of the logarithm. The only way to get a negative answer to an exponent is if the base is negative, and in logarithms the base is never negative. I'm not the best at explaining, but I'm pretty sure that's why -3 doesn't work.
-3 is a solution to the equation log6 x(x+1) = 1 but it is not a solution to the original equation. When we combined the logs using the product property, we lost the information that x must be >0 which came from log6 x. Whenever we lose information during some transformation, extraneous solutions show up and we have to check.