How do I find x? Exponential equation with two different bases. Reddit precalculus r/Homworkhelp

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#math #algebra #mathbasics

[daenerystargaryen]

I don't know why I'm watching this at 5AM since I'm a physicist doing PhD in neurophysics and computational neuroscience, but I thoroughly enjoyed this. 10/10. Younger generations are so lucky that they have someone like you explaining maths. Hopefully they'll know how to appreciate it and not waste their brains away on TikTok...

[reesetit]

I'm just impressed how you write with two different colors in one hand.

[keroro]

"How to find X?"

Bro, it's time to move on. Your X doesn't care about you anymore.

[MC-qmjn]

i took precalc 4 years ago and was arbitrarily recommended this video yet I still feel compelled to do the homework this man has given

[Gamert]

I'm in 10th grade, so whenever he says "let's use this rule" I'm just like "uh huh"
Edit: it's crazy how different some curriculums are in other countries.

[jordananderson]

To expand it, you use change of base to get log(96)/log(2/3). When dividing a logarithm, of course, you subtract the log of the denominator from the log of the numerator, which gives log(96)/(log2-log3). We can take the prime factors of 96: 3 and 2⁵, to get log(3•2⁵)/(log2-log3). With multiplication of logarithms, you add the logs of the multiplicands, so (log3 + log(2⁵))/(log2-log3). Finally, with exponentiation, you multiply the logarithm of the base by the exponent, which gives (log3 + 5log2)/(log2-log3).

[shrekyboi]

I solved it slightly different. I recognized that 3^(x+1) can be rewritten as [(1.5)(2)]^(x+1), which can be expanded as 1.5^(x+1) 2^(x+1). This is very helpful as it gives us an exponential of base 2 on both sides of the equation, which allows us to cancel out the x on the left side through exponent division rule. The full solution is below:

2^(x-5) = 3^(x+1)
2^(x-5) = [(1.5)(2)]^(x+1)
2^(x-5) = 1.5^(x+1) 2^(x+1)
2^(x-5)/2^(x+1) = 1.5^(x+1)
2^(-6) = 1.5^(x+1)

Now we only have a single x variable to deal with, so we could simply apply log to both sides and isolate for x

log[2^(-6)] = log[1.5^(x+1)]
(log[2^(-6)]/log[1.5]) - 1 = x
-11.257 = x

[jameshy]

australian here, i used my calculator. ive only seen the thumbnail and came straight here. the answer i got was (-ln(96))/ln(3/2) or approximately-11.257

edit: finished the video now and checked those two values of x. both were equal to my above answer. very nice 👍

[akifhossain]

I have a Zoology exam tomorrow. It's 3am. 10/10

[Dr.Insomniac.]

So I'm a 3rd year medical student watching this video and I dearly enjoyed it. Its like going down the memory lane. Really smooth teaching. Kudos to you..❤

[lucsas]

the second option is always what comes to my mind first, i find it way easier and more intuitive, but ive forced myself doing the natural base method too cuz you have to know them both imo

[sachinjain]

I Just Saw the Thumbnail And Thought " Ehhhh That looks Ez Lets Just Do It " Only to waste 30 mins And Find Out It Have Logarithm Which I Havent Studied😂

[brown_bread_]

I gave up on maths nearly 7 years ago in school. In my post graduation i watch this and feel my antipathy towards the subject reduce a little. Thanks

[oreivankovic]

bro fumbels my brain and proceedes to say:"but, here is a prettier way to do it"

[zuhakhalid]

That was a really good explanation! Thank you for explaining so clearly! 👏

[DA-gsgu]

I would consider simplify it with log to the base 10 which yields the same answer as the answer you obtained.
We could write it as,

X-5log(2)=X+1log(3)
Which on further simplification can provide,
x= -6.58/0.58= -11.3

And the answer you obtained at the end,

log (base)2/3 (96)= -11.26 (approx)
I feel its less hectic

[CuriousCyclist]

Thank you for taking the time to make this video. Much appreciated. ❤

[siddheshvispute]

Actually the equation becomes easy, when you use log in exponential problems.
Thanks ❤🇮🇳

[omverma_]

Bro I'm in 10th grade and I reached (2/3)^x = 96 and was like, "Now what?". Then I realised "Oh, this is out of bounds" 💀💀💀💀

[lukaskamin]

you can do backwords in 6:00 ONLY IF a and b are both positive (theoretically a can be 0, but it's a disputable question)