Help with AB question...

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<p><a href="http://www.collegeboard.com/prod_downloads/ap/students/calculus/ap06_sg_calculus_ab.pdf%5B/url%5D"&gt;http://www.collegeboard.com/prod_downloads/ap/students/calculus/ap06_sg_calculus_ab.pdf&lt;/a&gt;&lt;/p>

<p>In question 5b, how exactly did the equation get into a "y=" form? Also, whenever I have an equation with y as an absolute value, how would I algebraically deal with the brackets? Any quick response would be greatly appreciated.</p>

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<h1>? 10char.</h1>

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<p>Try again. The question number is there now.</p>

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<p>I think:

  1. x< 0 (so x = -1,-2, etc etc)</p>

<p>-2 times x will ALWAYS give you a positive # … thus the abs value is not needed.</p>

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<p>write it as (y+1)^-1 dy = x^-1 dx. Then take the anti-derivative to get the ln(1+y)=lnx + c. Finally, exponentiate and whatnot.</p>

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<p>You misunderstood what I wanted Hello, but I understand the algebra. There’s a part where I have to convert the natural logarithm into an exponent; and there’s a special rule when a logarithm is a power.</p>

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<p>Maybe someone can explain the second part of 5b and all of 5d? I’m trying to figure them out; and I can’t understand how they were able to get the values of m and b.</p>

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<p>Oh, so you mean 1+y = e^(ln|x| +c). e^c is a number, so e^c can be simply represented as C and multiplied by the exponent:Ce^(ln|x|), or c|x|.</p>