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C3 Maths help needed - functions :(

Not sure how I go about with 2b

2EC2EAFD-CC3B-4021-A314-D393E7A2EA74.jpg

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You were on the right track with 2e2x+3=82e^{2x+3} = 8. Solve this for x by first dividing both sides by 2. Ignore the restrictions for now, if they're just confusing you, it will be clear once you isolate xx.
Original post by Mystelle
Not sure how I go about with 2b



You started off fine. Find the x-coordinate where the gradient is 8, and rearrange into their wanted form.
Hmm. I think you need to look over the first question again first.
Original post by etothepiiplusone
Hmm. I think you need to look over the first question again first.


Looks like they've just drawn their axes strangely, they've labelled the asymptote correctly.
Reply 5
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frostfly
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Original post by RDKGames
You started off fine. Find the x-coordinate where the gradient is 8, and rearrange into their wanted form.


857B2E02-FEB8-4C0E-9B61-2FFF275F43D0.jpg

Is that how you do it? I’m not sure if, when dividing by 4, you divide the whole expression by 4 so including the ln or just the (8-6) like I have done
Original post by Mystelle


Is that how you do it? I’m not sure if, when dividing by 4, you divide the whole expression by 4 so including the ln or just the (8-6) like I have done


It's ln⁡(8)−64\frac{\ln(8)-6}{4}, not ln⁡(8−6)4\frac{\ln(8-6)}{4}
Reply 7
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frostfly
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Original post by RDKGames
It's ln⁡(8)−64\frac{\ln(8)-6}{4}, not ln⁡(8−6)4\frac{\ln(8-6)}{4}


03319DE7-24CB-4A2C-8A98-A2B66BE08818.jpg

Is it correct now?

*sorry for the sideways photos, no idea how it turned like that
Original post by Mystelle


Is it correct now?

*sorry for the sideways photos, no idea how it turned like that


Okay, hold on, back up - didn't notice it at first glance but you should have started with 2e2x−3=82e^{2x-3}=8 so e2x−3=4e^{2x-3}=4 hence 2x−3=ln⁡(4)2x-3=\ln(4)

Also ln⁡(b)a≠ln⁡(ba)\frac{\ln(b)}{a} \neq \ln(\frac{b}{a})
Reply 9
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frostfly
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Original post by RDKGames
Okay, hold on, back up - didn't notice it at first glance but you should have started with 2e2x−3=82e^{2x-3}=8 so e2x−3=4e^{2x-3}=4 hence 2x−3=ln⁡(4)2x-3=\ln(4)

Also ln⁡(b)a≠ln⁡(ba)\frac{\ln(b)}{a} \neq \ln(\frac{b}{a})


It says that the answer is a = 2, b = -1.5
but considering what you just said, that ln⁡(b)a≠ln⁡(ba)\frac{\ln(b)}{a} \neq \ln(\frac{b}{a}), I don't understand how they got their answer
Original post by Mystelle
It says that the answer is a = 2, b = -1.5
but considering what you just said, that ln⁡(b)a≠ln⁡(ba)\frac{\ln(b)}{a} \neq \ln(\frac{b}{a}), I don't understand how they got their answer


Carry on working from where I left off for you. Use the rule aln⁡(b)=ln⁡(ba)a\ln(b)=\ln(b^a)
Reply 11
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frostfly
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Original post by RDKGames
Carry on working from where I left off for you. Use the rule aln⁡(b)=ln⁡(ba)a\ln(b)=\ln(b^a)


Cheers, got there in the end :smile:

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