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    Not sure how I go about with 2b

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    You were on the right track with 2e^{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 x.
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    (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.
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    Hmm. I think you need to look over the first question again first.
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    (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.
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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.
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    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
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    (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 \frac{\ln(8)-6}{4}, not \frac{\ln(8-6)}{4}
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    (Original post by RDKGames)
    It's \frac{\ln(8)-6}{4}, not \frac{\ln(8-6)}{4}
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    Is it correct now?

    *sorry for the sideways photos, no idea how it turned like that
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    (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 2e^{2x-3}=8 so e^{2x-3}=4 hence 2x-3=\ln(4)

    Also \frac{\ln(b)}{a} \neq \ln(\frac{b}{a})
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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 2e^{2x-3}=8 so e^{2x-3}=4 hence 2x-3=\ln(4)

    Also \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 \frac{\ln(b)}{a} \neq \ln(\frac{b}{a}), I don't understand how they got their answer
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    (Original post by Mystelle)
    It says that the answer is a = 2, b = -1.5
    but considering what you just said, that \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 a\ln(b)=\ln(b^a)
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    (Original post by RDKGames)
    Carry on working from where I left off for you. Use the rule a\ln(b)=\ln(b^a)
    Cheers, got there in the end

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