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Algebra

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Solve the equation.
2log3 y = log3 4+ log3 (y+8)

Here's what ive got
take out the logs and say y^2= 4+ y +8
which simplifies to (y^2)-y=12
how do i simplfy the y?

  • Algebra - ,

    note that you cannot simplify the right side equation log3 (4) + log3 (y+8) , by 4 + y + 8 ... they should be multiplied. recall some laws of exponents; if we are add terms of the same base, for example
    log 3 + log 5
    this simplifies to
    log(3*5) or log 15
    going back to the problem, the equation simplifies to
    2log3 y = log3 4 + log3 (y+8)
    log3 (y^2) = log3 (4*(y+8))
    log3 (y^2) = log3 (4y + 32)
    cancelling out the log3,
    y^2 = 4y + 32
    y^2 - 4y - 32 = 0
    this is a quadratic equation. since it's factorable, let's just factor this one:
    (y - 8)(y + 4) = 0
    y = 8 and y = -4
    note that y = -4 is extraneous, since if we substitute this back to the original, and take its log:
    left side: 2log3 (-4) = undefined
    thus,
    y = 8

    hope this helps~ :)

  • Algebra - ,

    Jai, your a life saver! you explained everything perfectly! thank you sooo much!

    that helped a ton, thanks! : )

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