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Meehan98

  • one year ago

So, I need some help with Logarithms. The equation is: log (subscript 3) x^2 + log (subscript 3) 9 = 4. I thought you multiply (x^2)(9) in order to get one log on the left side, so you're left with... log (subscript 3) 9x^2=4 and because the log is the exponent, you have log (subscript 3) 9x^2=3^4=81, which makes x=3, but that is the wrong answer. Thank You!

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  1. freckles
    • one year ago
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    \[\log_3(x^2)+\log_3(9)=4 \\ \text{ yes you can use product rule } \\ \log_3(9x^2)=4\]

  2. freckles
    • one year ago
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    you can then write in the equivalent exponential form

  3. freckles
    • one year ago
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    \[\log_b(x)=y \implies b^y=x\]

  4. freckles
    • one year ago
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    \[3^{4}=9x^2\]

  5. freckles
    • one year ago
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    ok and I think you said that above

  6. freckles
    • one year ago
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    \[81=9x^2 \\ \frac{81}{9}=x^2 \\ x^2=9\] you should get two solutions not just one

  7. freckles
    • one year ago
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    x=3 is right but there is one more solution

  8. Meehan98
    • one year ago
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    Oh, so it would be 3, -3.

  9. freckles
    • one year ago
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    yep

  10. Meehan98
    • one year ago
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    Ok, thank you! For some reason, I had the mindset that it couldn't be a negative number. You helped so much!

  11. freckles
    • one year ago
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    when you plug in -3 into the equation you have that one term is log((-3)^2)) the inside of that log is positive so it is cool

  12. freckles
    • one year ago
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    you know since (-3)^2 =9 which is a positive number

  13. freckles
    • one year ago
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    your aim is to make sure the solutions you get do not make the inside of the logs negative or zero

  14. Meehan98
    • one year ago
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    Okay, that makes sense! Thank you!!

  15. freckles
    • one year ago
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    \[f(x)=\log(x^2) \text{ has domain all real numbers } \\ \text{ while } g(x)=\log(x) \text{ has domain } x>0\]

  16. freckles
    • one year ago
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    so if the equation was: \[2 \log_3(x)+\log_3(9)=4 \\ \text{ you would want \to make sure your solutions satisfy } x>0\]

  17. freckles
    • one year ago
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    \[\log_3(x^2)+\log_3(9)=4 \text{ by power rule } \\ \log_3(9x^2)=4 \text{ by product rule } \\ 3^{4}=9x^2 \text{ equivalent exponential form } \\ 81=9x^2 \\ x^2=9 \\ x=\pm 3 \\ \text{ but we said } x>0 \text{ so only } x=3 \\ \text{ works for } 2 \log_3(x)+\log_3(9)=4\]

  18. freckles
    • one year ago
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    I made a mistake above with the domain of f

  19. freckles
    • one year ago
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    \[f(x)=\log(x^2) \text{ has domain all real numbers excluding } x=0\]

  20. freckles
    • one year ago
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    anyways I hope you see I was solving a slightly different equation right there

  21. freckles
    • one year ago
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    \[2 \log_3(x)+\log_3(9)=4 \text{ has solutions } x=3 \\ \log_3(x^2)+\log_3(9) =4 \text{ has solutions } x=3,-3\]

  22. Meehan98
    • one year ago
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    Okay, you have made this a lot easier! The pieces of information finally fit now!! Thank you so much!

  23. freckles
    • one year ago
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    np

  24. freckles
    • one year ago
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    I gave you medal because you are awesome. You showed your work. And that totally rules in my opinion. You also did pretty well on solving it.

  25. Meehan98
    • one year ago
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    Thanks! I'm pretty new to this site, and I will definitely stick with it because the help is awesome!

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