@ChelseyKryou, I assume you are referring to the step at 0:57. You are correct that (x^2)(x^3) = x^5, but in the example at 0:57 we are simplifying (2^3)^(2x+1) and in this case when you take a power to a power the exponents are multiplyed. The general rule looks like this: (x^a)^b = x^(ab). So (x^2)(x^3) = x^5 and (x^2)^3 = x^6. I hope that helps.
You are doing great so far. Just to clarify, the original problem is 2^(x+3) = 3^(x-4). Your next step is to divide both sides by log3 - log2. As far as how to write the final answer it depends what the directions say. If they say to round to the nearest whatever, you can plug your answer into the calculator. If it says to give the exact answer you will want to simplify things using the rules of logs. For example 4log3 => log3^4 => log81, log3 - log2 => log1.5, also simplify the sum of logs.
Wow, that's an interesting way to do the problem. It does work. If this method is easier for you I suggest you use it. My theory is that the easiest method is the one that you understand the best. Your method requires a few substitution steps that some might struggle with. I really appreciate seeing different approaches. Thanks for sharing!
Thank you for the help man, studying for a final right now and gotta say, you going over the laws of exponents at the beginning really helped it click.
Hey man, I understand what you're doing and all and did great in my Homework until I came to this problem: 2^x+3=3^x-4 I did what you said and got this: (x+3)log2=(x-4)log3 then got xlog2+3log2=xlog3-4log3 I moved them to like sides and got: xlog3-xlog2=4log3+3log2 Then got: x(log3-log2)=4log3+3log2 What do I do on the right side, and how would I get the answer to this problem?
Nice video. I found a different way to calculate it that is a lot easier. Just like on your example of that equation with similar bases(2^3X=8^2x+1), we can use log to calculate 9^X in the base of 5: 5^Y = 9 and Y=Log5to9 So Y = Log9/Log5 (To pass it all to the base of 10) So, back at the original equation: 5^(x+2) = 5^(Log9/Log5)^x, which brings you to: X+2 = (Log9/Log5)^X, which can easily be calculated using a calculator, since it's all in the base of 10.
I was wondering if you could help me with a similar but more difficult problem. It is as shown below: 4^x times 5^4x+3 = 10^2x+3 Any tips would be appreciated!
Good video. You do not need a parenthesis around a product when dividing, eg (2log5)/log(9/5)=2log5/log(9/5). You would need a parenthesis if you had more than one term in the numerator.
I would give my understanding of this video a 3/4 because I understand why he made them both logs because he needed the same base of 10 but when he isolated the x it was kinda confusing so I wish he solved one more problem since I’m still kinda unfamiliar with it.
I can do a video with natural logs on both sides. Would you like to a problem like 1) ln (something) = ln(something) or 2) ln(something) = ln (something) + #.
Couldn't find this on Khan Academy! Thanks so much!
I've been searching for hours on how to do this correctly. Thank you so much; it really helped.
@ChelseyKryou, I assume you are referring to the step at 0:57. You are correct that (x^2)(x^3) = x^5, but in the example at 0:57 we are simplifying (2^3)^(2x+1) and in this case when you take a power to a power the exponents are multiplyed. The general rule looks like this: (x^a)^b = x^(ab). So (x^2)(x^3) = x^5 and (x^2)^3 = x^6. I hope that helps.
You helped me so, much I could just give you a huge hug. Thank you
Funnny
You are a saint! I've been trying to figure this out for about a week and then i found this video. You saved my life XD
So glad you found the video and it helped!
You are doing great so far. Just to clarify, the original problem is 2^(x+3) = 3^(x-4). Your next step is to divide both sides by log3 - log2. As far as how to write the final answer it depends what the directions say. If they say to round to the nearest whatever, you can plug your answer into the calculator. If it says to give the exact answer you will want to simplify things using the rules of logs. For example 4log3 => log3^4 => log81, log3 - log2 => log1.5, also simplify the sum of logs.
Wow, that's an interesting way to do the problem. It does work. If this method is easier for you I suggest you use it. My theory is that the easiest method is the one that you understand the best. Your method requires a few substitution steps that some might struggle with. I really appreciate seeing different approaches. Thanks for sharing!
Thank you for the help man, studying for a final right now and gotta say, you going over the laws of exponents at the beginning really helped it click.
Excellent video; very thorough and clear, and you explained why you did this and not that. Thank you so much.
great job dude. this was so much more efficient than any of the times i tried to learn this in classes
holy shit thank you i have been looking all through these damn videos and could never fing the exact approach i needed at 3:39 again thank you!!!!
Hey man, I understand what you're doing and all and did great in my Homework until I came to this problem:
2^x+3=3^x-4
I did what you said and got this: (x+3)log2=(x-4)log3
then got xlog2+3log2=xlog3-4log3
I moved them to like sides and got: xlog3-xlog2=4log3+3log2
Then got: x(log3-log2)=4log3+3log2
What do I do on the right side, and how would I get the answer to this problem?
Nice video. I found a different way to calculate it that is a lot easier.
Just like on your example of that equation with similar bases(2^3X=8^2x+1), we can use log to calculate 9^X in the base of 5:
5^Y = 9 and Y=Log5to9
So Y = Log9/Log5 (To pass it all to the base of 10)
So, back at the original equation:
5^(x+2) = 5^(Log9/Log5)^x, which brings you to:
X+2 = (Log9/Log5)^X, which can easily be calculated using a calculator, since it's all in the base of 10.
fantastic explanation - Looked for a long time and this is awesome
@rebym same idea. Use logs, there's just an extra step I think
I was wondering if you could help me with a similar but more difficult problem. It is as shown below:
4^x times 5^4x+3 = 10^2x+3
Any tips would be appreciated!
x^2 times x^3 doesnt equal x^6. It equals x^5. Since you use addition when it comes to exponential multiplicational integrers.
Good video. You do not need a parenthesis around a product when dividing, eg (2log5)/log(9/5)=2log5/log(9/5). You would need a parenthesis if you had more than one term in the numerator.
How would you solve if the problem was (2^x)+3=(3^x)-4
Thanks man! I feel like this helped me more than my teacher haha
thanks for the great explanation!
+Daniel Chen you are welcome!
I would give my understanding of this video a 3/4 because I understand why he made them both logs because he needed the same base of 10 but when he isolated the x it was kinda confusing so I wish he solved one more problem since I’m still kinda unfamiliar with it.
thanks im still taking geometry but i get this cos you made it easy
thanks! you teach well! your video helped me a lot :^)
this was very helpful! thank you!!
Why do none of these tutorials have anything on natural logs on both sides?
I can do a video with natural logs on both sides. Would you like to a problem like 1) ln (something) = ln(something) or 2) ln(something) = ln (something) + #.
yo, I saw in a video where if it is for example 2log5, the 2 goes to the exponent and its equal to log1/25. U forgot that step
Please note I do not want the answer, I will figure that out myself, I just want to know what step comes after this.
Omg ! this help me a lot ! Thank u
you are the best!!
i learned more than i did in 5 years of schooling...lol
this was super helpful, thanks
log(5/9) would give you a negative, and therefore your answer would be negative, correct?
Yes log of (5/9) would be a negative number.
THANK YOU!!! so helpful!!
YOU SAVED MY LIFE
Thank you so much
Thank you sooooo much!!:)
Excellent, glad it helped!
thank you so much man
Thank you very much.
thank you sooooo much!
did not undertsand anything. help
thanks bro.
THANK YOU!
Okay, thanks, man :)
bless you. thank you so so much! :)
hoho
thanks
was
blah blah*blah=blah
thank you so much!