I found this site with online problems and answers. https://courses.lumenlearning.com/waymakercollegealgebra/chapter/expand-and-condense-logarithms/

I've tried several problems and my answer is always wrong.

I've added two screenshots: 1) the problem and the answer according to the site

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2) my work and answer

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Could someone please explain what I'm doing wrong? I don't understand why the denominator disappears.


  • $\begingroup$ You should have $\frac{1}{2}\ln (y + 7) - \ln (y + 7) = -\frac{1}{2} \ln (y + 7)$ in your final step. $\endgroup$ – N. F. Taussig Aug 4 '18 at 7:42
  • $\begingroup$ Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$. $\endgroup$ – gimusi Aug 4 '18 at 8:00
  • $\begingroup$ @LiesVanRompaey Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Sep 6 '18 at 23:02

Because the domain gives $y>7$ and $$\ln\frac{\sqrt{(y^2-49)(y+8)^6}}{y+7}=\ln\frac{\sqrt{y-7}|y+8|^3}{\sqrt{y+7}}=\frac{1}{2}\ln(y-7)-\frac{1}{2}\ln(y+7)+3\ln|y+8|=$$ $$=\frac{1}{2}\ln(y-7)-\frac{1}{2}\ln(y+7)+3\ln(y+8).$$

  • 1
    $\begingroup$ I thought exactly to the same explanation...the key point indeed is that $x>7$ :) $\endgroup$ – gimusi Aug 4 '18 at 7:58

Assuming for the definition of the given expression that $y>7$ we can avoid absolute values when extracting from square roots since all factors are positive and your result appears to be correct.


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