Eris13696
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I hope everyone is doing good today. I understand how to do this question and all the steps involved in it. But however I don’t get how to simplify it like they did here. (If I want to click on explain more I have to pay).
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davros
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(Original post by Eris13696)
I hope everyone is doing good today. I understand how to do this question and all the steps involved in it. But however I don’t get how to simplify it like they did here. (If I want to click on explain more I have to pay).
which bit is confusing you - these are just standard log rules so

2ln(3) = ln(3^2) = ln 9
2xln3 = ln(3^(2x)) etc
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Eris13696
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(Original post by davros)
which bit is confusing you - these are just standard log rules so

2ln(3) = ln(3^2) = ln 9
2xln3 = ln(3^(2x)) etc
What’s confusing me is how the simplify it, this is how I do it. This is the answer I get without using their simplification. If you put values in both mine and theirs, you get the same thing. If I kept this in the exam, will it be fine or do I have to always simplify?
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davros
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(Original post by Eris13696)
What’s confusing me is how the simplify it, this is how I do it. This is the answer I get without using their simplification. If you put values in both mine and theirs, you get the same thing. If I kept this in the exam, will it be fine or do I have to always simplify?
Well, your answer isn't technically wrong (as far as I can see from a quick glance) but you've made an awful lot of extra work for yourself!

e^{-2x\ln 3} is just of the form e^{kx} with k = -2ln3, so you should be able to integrate this straight off to get:

\dfrac{1}{k}e^{kx} as it is a standard integral.

You can then choose whether to write the solution as a power of e (which is how the original integrand looks), or convert it to a power of 3 or a power of 9. Personally I don't see anything to choose between 3^{2x} and 9^x
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Eris13696
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(Original post by davros)
Well, your answer isn't technically wrong (as far as I can see from a quick glance) but you've made an awful lot of extra work for yourself!

e^{-2x\ln 3} is just of the form e^{kx} with k = -2ln3, so you should be able to integrate this straight off to get:

\dfrac{1}{k}e^{kx} as it is a standard integral.

You can then choose whether to write the solution as a power of e (which is how the original integrand looks), or convert it to a power of 3 or a power of 9. Personally I don't see anything to choose between 3^{2x} and 9^x
I’m so grateful that you answered me, you’re the best one at simplifying math here 😭 I spent 2 hours yesterday trying to understand how it’s done and then I found these steps on the app so I just went with whatever I had. Your method is done in like 2 steps compared to the one I used, thank you <3
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