Expanding log problem

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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



enter image description here



2) my work and answer



enter image description here



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



Thanks




algebra-precalculus proof-verification logarithms radicals absolute-value




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  • You should have $frac12ln (y + 7) - ln (y + 7) = -frac12 ln (y + 7)$ in your final step.
    – N. F. Taussig
    2 days ago










  • Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$.
    – gimusi
    2 days ago














up vote
0
down vote

favorite
1












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



enter image description here



2) my work and answer



enter image description here



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



Thanks




algebra-precalculus proof-verification logarithms radicals absolute-value




share|cite|improve this question





















  • You should have $frac12ln (y + 7) - ln (y + 7) = -frac12 ln (y + 7)$ in your final step.
    – N. F. Taussig
    2 days ago










  • Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$.
    – gimusi
    2 days ago












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





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



enter image description here



2) my work and answer



enter image description here



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



Thanks




algebra-precalculus proof-verification logarithms radicals absolute-value




share|cite|improve this question













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



enter image description here



2) my work and answer



enter image description here



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



Thanks





algebra-precalculus proof-verification logarithms radicals absolute-value





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share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Michael Rozenberg

86.9k1576178




86.9k1576178









asked 2 days ago









Lies Van Rompaey

1




1











  • You should have $frac12ln (y + 7) - ln (y + 7) = -frac12 ln (y + 7)$ in your final step.
    – N. F. Taussig
    2 days ago










  • Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$.
    – gimusi
    2 days ago
















  • You should have $frac12ln (y + 7) - ln (y + 7) = -frac12 ln (y + 7)$ in your final step.
    – N. F. Taussig
    2 days ago










  • Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$.
    – gimusi
    2 days ago















You should have $frac12ln (y + 7) - ln (y + 7) = -frac12 ln (y + 7)$ in your final step.
– N. F. Taussig
2 days ago




You should have $frac12ln (y + 7) - ln (y + 7) = -frac12 ln (y + 7)$ in your final step.
– N. F. Taussig
2 days ago












Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$.
– gimusi
2 days ago




Your solution is correct. It seems there is a typo in your source. As noticed it is important recognize that we are implicitly assuming $x>7$.
– gimusi
2 days ago










2 Answers
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Because the domain gives $y>7$ and
$$lnfracsqrt(y^2-49)(y+8)^6y+7=lnfracsqrty-7sqrty+7=frac12ln(y-7)-frac12ln(y+7)+3ln|y+8|=$$
$$=frac12ln(y-7)-frac12ln(y+7)+3ln(y+8).$$






share|cite|improve this answer

















  • 1




    I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
    – gimusi
    2 days ago

















up vote
0
down vote













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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    2 Answers
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    2 Answers
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    up vote
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    Because the domain gives $y>7$ and
    $$lnfracsqrt(y^2-49)(y+8)^6y+7=lnfracsqrty-7sqrty+7=frac12ln(y-7)-frac12ln(y+7)+3ln|y+8|=$$
    $$=frac12ln(y-7)-frac12ln(y+7)+3ln(y+8).$$






    share|cite|improve this answer

















    • 1




      I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
      – gimusi
      2 days ago














    up vote
    2
    down vote













    Because the domain gives $y>7$ and
    $$lnfracsqrt(y^2-49)(y+8)^6y+7=lnfracsqrty-7sqrty+7=frac12ln(y-7)-frac12ln(y+7)+3ln|y+8|=$$
    $$=frac12ln(y-7)-frac12ln(y+7)+3ln(y+8).$$






    share|cite|improve this answer

















    • 1




      I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
      – gimusi
      2 days ago












    up vote
    2
    down vote










    up vote
    2
    down vote









    Because the domain gives $y>7$ and
    $$lnfracsqrt(y^2-49)(y+8)^6y+7=lnfracsqrty-7sqrty+7=frac12ln(y-7)-frac12ln(y+7)+3ln|y+8|=$$
    $$=frac12ln(y-7)-frac12ln(y+7)+3ln(y+8).$$






    share|cite|improve this answer













    Because the domain gives $y>7$ and
    $$lnfracsqrt(y^2-49)(y+8)^6y+7=lnfracsqrty-7sqrty+7=frac12ln(y-7)-frac12ln(y+7)+3ln|y+8|=$$
    $$=frac12ln(y-7)-frac12ln(y+7)+3ln(y+8).$$







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered 2 days ago









    Michael Rozenberg

    86.9k1576178




    86.9k1576178







    • 1




      I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
      – gimusi
      2 days ago












    • 1




      I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
      – gimusi
      2 days ago







    1




    1




    I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
    – gimusi
    2 days ago




    I thought exactly to the same explanation...the key point indeed is that $x>7$ :)
    – gimusi
    2 days ago










    up vote
    0
    down vote













    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.






    share|cite|improve this answer

























      up vote
      0
      down vote













      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.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        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.






        share|cite|improve this answer













        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.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        gimusi

        63.6k73480




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