Solving & checking logarithmic equations (4 examples)

0:00 Q1 log_3(5-x)=4
1:39 Q2 log_2(x^2-7x)=3
3:54 Q3 log_9(x-5)+log_9(x+3)=1
6:14 Q4 log_3(10x+3)-log_3(2x-3)=2

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[rylegaming]

this youtube channel is my fav. channel because calculus is my fav. topic in mathematics . It makes me watch everyday

[wristdisabledwriter]

I was hoping you would do a question where one of the possible x is positive but not an answer because it ends up negative in the log

[alberteinstein]

These were very fun 😁🥳

I love these nice refreshers

[deveshswami]

As usual I was not told to raise to power of 3 on both, instead I was told to bring that 3 and raise it to the power 4, and that's Really confusing when base has a variable in it. And as always Thanks prof 😊👍🏻

[fizixx]

Prob one of the best applications using logs w/ different bases.

[BlaccSenpai]

What do you do when the logs have different bases?

[fabiotiburzi]

6:12 but in the immaginary world -4 give in the first equation -9 and that log is equal to i, so is legit

[Firefly]

If we get log(negative number), do we need to go to complex world to check if imaginary part cancel out?

[Daniel_Frise]

Obviously this channel is for high school calculus, but once you take complex analysis it’s hard to not consider such questions. I notice that if we were using the complex logarithm, then x=-4 would be a solution to the third equation. If we are working with the complex logarithm, then is it still possible to obtain extraneous solutions?

[reidpattis]

If the LAST problem had different coprime bases, say log base 3 and log base 5. I’m actually not too sure what to do.

[dqrksun]

Why can't you put negatives in the log?

[nidhiagrawal]

6:09 I disagree because log is also defined for negative numbers, it's just a non-real number. log(-x) = log(x) + (2n + 1)πi where n is in the set of integers.
Then log base 9(-9) + log base 9(-1) = 1 can be a true equation.
log base 9(-9) + log base 9(-1) = 1
log base 9(9) + (2n + 1)πi + log base 9(1) + (2m + 1)πi = 1 where n, m are in the set of integers
1 + 0 + ((2n + 1) + (2m + 1))πi = 1
(2n + 2m +2)πi = 0
2(n + m + 1) = 0
n + m + 1 = 0

So when n + m = -1 then x = -4 and n + m can be -1 when n = -m - 1 or m = -n - 1.
E.g., n = 0, m = -1.

So yes x = -4 is a solution.

Your mistake was assuming log(non-positive) is not defined but actually it is. You can see it by using e^x (its inverse). e^(2n + 1)πi gives a negative result where n is in the set of integers.

I have seen this mistake in a lot of your videos on log. Please improve on this. I don't want misconception to be spread around the world. ☺️