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Solving Exponential and Logarithmic Equations (Multiple Examples)
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- Опубліковано 24 лис 2019
- Learn how to solve both exponential and logarithmic equations in this video by Mario's Math Tutoring. We discuss lots of different examples such as the one to one property of exponents, one to one property of logs, changing forms from exponential to logarithmic form and vice versa, and we also discuss a problem involving factoring. We discuss how to recognize an extraneous solution as well as how to use the condensing properties for logs in some of the examples.
Related Videos to Help You Succeed!:
Logs - Everything You Need to Know
• Logs Everything You Ne...
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Mario, I have never seen a teacher who puts all the plausible cases that confuse students or make them to perform common illegal operation.Thank you .
Bee Bless, this video is not an introduction to logarithm, Mario has many introductory videos on this and many other subjects. These video is a comprehensive review of them and these examples are merely to clarify some common mistakes that students make.
You’re welcome. Glad you liked the video.
Keep going ur lectures were truly understandable🙏
Thanks you so much for this. All your videos help me with my math.
Truly amazing work thank you helped so much!!!!😃
I'm so glad for your presence on the Internet!
Mr. Mario thank you so much for your help, you saved me last unit man!🐐
This guy....✒✅💯💯... Quality Stuff.
thank you sir.
Thank you!!
Thank you!
Best Mario! it couldn't be easier.
Thank You!
Cheers mate, thanks for the video
Any time!
King you're amazing
Amazing!
Glad you liked the video Sheri!
This is the one person wo saved me on my tests
Glad to hear my videos helped you.
Bro helped me get into upenn ☠️ thank you
I looked at the thumbnail and I had no idea what it was until you explained it thx
Glad you liked my explanation!
thank you so much
You're welcome!
great video!
Thanks!
Lifesaver
Only God can reward Mario.
Greetings Sir, This video I like, but I feel it needs a little more explanation in a few steps. I will continue to review to keep up. I don't get the natural log, log, and exponential aspect and what all means. I just see just do it.
This video I did here may help you to understand logarithms better: Logs Everything You Need to Know ua-cam.com/video/LCmnKWXCIfk/v-deo.html
I feel like I can watch these all day and still not understand logarithms. idk what it is about them but they hurt my brain.
Watching these vids cuz of a test i have on monday and seems as the teacher is starting to not like 💀so wish me luck guys 😭🤞🏻
Test next period, let’s see
I d8dnt know that maths is like a spreading skill or a bird eye view skill.
Am I the only one triggered? 5:15
Ok.
If it helps, first think of that equation as this: (e^x)^2-(e^x)-12=0.
Remember that there is a property of exponents that says that if you have a power raised to a power, you leave the base alone, and then multiply the exponents together (However, here, you're using this property, the other way around). Therefore, e^(2x)=(e^2)^x=(e^x)^2 (More preferably, for the sake of this problem). Then, the middle e^x is just (e^x)^1. Hence, we have a quadratic equation in terms of e^x. The strategy that I (personally) use is to make a substitution of variables. I am going to let u=e^x (I could have used any different variable, but I use u, as it's conventional notation).
This equation now becomes u^2-u-12=0. Now, we solve for u. We can factor u^2-u-12 as (u+3)*(u-4). Hence, (u+3)*(u-4)=0, and now we use the Zero Product Property/Theorem, set each separate factor equal to zero (0), and solve for u.
u+3=0 => u=-3 (Subtract 3 from both sides)
u-4=0 => u=4 (Add 4 to both sides)
Our original equation, though, was in terms of x, not in terms of u, so we need to switch back to x's. Remember that u=e^x. Hence, u=-3 becomes e^x=-3 (This is not possible, however, because e^x is always positive, for all real values of x, so this equation has no solution (e^x is NEVER equal to zero (0) or a negative number), and we can move on to the other equation) and u=4 becomes e^x=4 (4 is a positive number, but since e is not a power of 4, and 4 is not a power of e, we have to convert this equation to logarithmic form), so log base (e) of (4)=x. Finally, recall that log base (e) is equivalent to ln. Therefore, ln(4)=x, or, in other words, x=ln(4), which is the only solution to this equation.
Can you explain a little bit slower, it's hard to follow sometimes! 😅
Noted!
If you want a one on one tutoring, he gonna charge you . 😂
Although that I’m not an English native speaker I found it easy to understand the lecture, Thanks a lot 🤍
Glad to hear that!