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Implicit equations help

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Original post by SDavis123
Yh I did, did you see the picture?


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I hadn't but I have now.

So make y the subject. There are many ways you can do it. Combining the logarithms would be a sensible step and then you are going to need to raise both sides to base e.
Reply 21
Original post by Mr M
I hadn't but I have now.

So make y the subject. There are many ways you can do it. Combining the logarithms would be a sensible step and then you are going to need to raise both sides to base e.


If before was painful, I think what your about to see might just be enough to knock you out! :tongue:




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Original post by SDavis123
If before was painful, I think what your about to see might just be enough to knock you out! :tongue:




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You have some really worrying misconceptions.

a2+b2a+b\sqrt{a^2+b^2} \neq a + b
Reply 23
Original post by Mr M
You have some really worrying misconceptions.

a2+b2a+b\sqrt{a^2+b^2} \neq a + b


I'm not getting anywhere, sorry for the headache that I've caused you and thanks for trying to help :smile:


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Original post by SDavis123
I'm not getting anywhere, sorry for the headache that I've caused you and thanks for trying to help :smile:


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Multiply through by 2 and combine the logarithms. Now exponentiate and you are pretty much there.
Reply 25
12ln(y2+1)ln(y+1)\frac{1}{2}\ln{(y^{2} + 1)} \not= \ln{(y + 1)}

^ Referring to what Mr M was saying.


Original post by Mr M
a2+b2a+b\sqrt{a^2+b^2} \neq a + b

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