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xBeEnchantedx

  • 6 months ago

How do you solve: logbase5 (square root of x+3) = logbase5 (4)?

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  1. myininaya
    • 6 months ago
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    this means your insides have to be equal

  2. xBeEnchantedx
    • 6 months ago
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    \[\log_{5} (\sqrt{x+3})=\log_{5} (4)\] <- thats a better way to view the equation.

  3. myininaya
    • 6 months ago
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    Since y=log_5(x) is a one to one function if you have f(a)=f(b) then this implies a=b

  4. myininaya
    • 6 months ago
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    like basically i'm telling you that \[\sqrt{x+3}=4 \text{ now solve }\]

  5. xBeEnchantedx
    • 6 months ago
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    Ohhh thank you. I accidentally put the wrong exponent when trying to get rid of the square root, which through off my calculations. Thank you :D

  6. myininaya
    • 6 months ago
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    coolness

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