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Mech88

  • 2 years ago

2logx to the base 3 - 3log4 to the base 2 = log1 to the base b

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  1. SolomonZelman
    • 2 years ago
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    What is the question?

  2. SolomonZelman
    • 2 years ago
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    Oh the b?

  3. Mech88
    • 2 years ago
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    Solve for x

  4. campbell_st
    • 2 years ago
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    is this the question? \[2\log_{3}(x) - 3\log_{2}(4) = \log_{b}(1)\]

  5. Mech88
    • 2 years ago
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    Yes... That's the question thank you

  6. campbell_st
    • 2 years ago
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    ok... so \[\log_{2}(4) = \log_{2}(2^2) = 2\] so you have \[\log_{3}(x^2) - 6 = \log_{b}(1)\] so I'd suggest using change of base...

  7. Mech88
    • 2 years ago
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    How would I do that? I also get to that step but can't get further

  8. campbell_st
    • 2 years ago
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    ok... so this is how I would do the question, after re-reading it the log of 1 to any base is always zero... its just a fact. and you saw how I managed the base 2 log of 4 so you have \[2\log_{3}(x) - 6 = 0\] which becomes \[2 \log_{3}(x) = 6\] divide both sides of the equation by 2 \[\log_{3}(x) = 3\] now raise each side of the equation to the power of 3 \[3^{\log_{3}(x)} = 3^3\] which is x = 27

  9. Mech88
    • 2 years ago
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    Thank you so much

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