hello for 3:37, why can't I just simply multiply by e on both side? and get x=2e. why do i have to power it in terms of e? thank you very much for this video!
@@BicenMaths ohhhh i got it now! i thought lne^x written at the beginning meant that they are multiplied. But actually it is e^x in the ln. Thank you for clarifying!
If you log equations, you have to log the entire side of the equation, not the terms separately (a bit like if you have 2 + 3 = 5, you can’t square each part individually, otherwise you get 4 + 9 = 25 which obviously isn’t true). You could log the whole thing, but you’d get stuck. Instead, in this video I’m wanting you to recognise that these equations look like quadratics, and can’t be solved (at least to start with!) using logs or indices.
clear explanation, thank you.
thank so much sir. the way u teach is better than maths genie and tlmaths. AS maths isn't looking so intimidating now. cheers
Pleased to hear you're not feeling so intimidated! This is a tough chapter, well done!
hello for 3:37, why can't I just simply multiply by e on both side? and get x=2e. why do i have to power it in terms of e? thank you very much for this video!
Because the 'e' and the 'ln' only cancel if the ln is in the power of the e. Not just being multiplied together! Hope that makes sense!
@@BicenMaths ohhhh i got it now! i thought lne^x written at the beginning meant that they are multiplied. But actually it is e^x in the ln. Thank you for clarifying!
At 4:00 why wouldnt 5 by Ln5?
Hmmm I am not sure I understand your question, sorry!
4:20 why does the e^ln cancel out?
This is the definition of exponentials and logarithms and how they behave!
Inverses of each other
8:10 why can't you log the constants?
If you log equations, you have to log the entire side of the equation, not the terms separately (a bit like if you have 2 + 3 = 5, you can’t square each part individually, otherwise you get 4 + 9 = 25 which obviously isn’t true). You could log the whole thing, but you’d get stuck. Instead, in this video I’m wanting you to recognise that these equations look like quadratics, and can’t be solved (at least to start with!) using logs or indices.
@@BicenMaths ohhhh, I understood it as soon as I read ur first sentence. Thanks a lot!!!