log₃(x+5)+log₃(x−7)=2, my answer is in the description please check it out?

Correct me if I'm wrong ^_+

My answer is "x =1+3√5", because it says decimal it becomes "7.7" right?

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  • Philip
    Lv 6
    1 month ago

    Notation: Put log,base3 = L.;

    L(x+5) + L(x-7) = 2, ie., L((x+5)(x-7)) = 2, ie., 3^L((x+5)(x-7)) = (x+5)(x-7) = 3^2.

    (x+5)(x-7) = 9, ie., x^2 -2x -44 = 0. Then 2x = 2(+/-)D, where D^2 = 4 + 4*44 =;

    4*45= 4*9*5 = (6^2)(rt5)^2 = (6rt5)^2. Then x = 1 + 3rt5, rt = ''square root of''.;

    Negative root 1-3rt5 is invalid for x since both (x+5) & (x-7) must be positive.;

    x = 7.708203933 = 7.7 rounded off to 1 decimal place.

     

  • 1 month ago

    log₃(x + 5) + log₃(x - 7) = 2

    so, log₃[(x + 5)(x - 7)] = 2  

    Hence, (x + 5)(x - 7) = 3²

    i.e. x² - 2x - 44 = 0

    Using the quadratic formula we get:

    x = 1 ± 3√5  

    Now, x = 1 + 3√5 is the only valid solution as x + 5 > 0 and x - 7 > 0

    so, x = 7.7...to 1 d.p.

    :)>

  • 1 month ago

    log₃(x + 5) + log₃(x − 7) = 2

    log₃[(x + 5)(x − 7)] = 2

    (x + 5)(x − 7) = 3²

    x² - 2x - 35 = 9

    x² - 2x - 44 = 0

    x = {-(-2) ± √[(-2)² - 4(0)(-44)]} / [2(1)]

    x = (2 ± √180)/2

    x = 1 + 3√5  or  x = 1 - 3√5 (rejected for x - 7 < 0)

    Hence, x = 1 + 3√5 ≈ 7.7

    Your answer is correct.

  • 1 month ago

    log₃(x + 5) + log₃(x - 7) =2

    Use product rule:

    log₃[(x + 5)(x-7)] = 2

    Exponentiate using 3 on both sides:

    (x + 5)(x - 7) = 3^2

    (x + 5)(x - 7) = 9

    x^2 - 2x - 35 = 9

    x^2 - 2x - 44 = 0

    Using quadratic formula, the roots are 7.708203932 and -5.708203932.

    But, the negative answer is extraneous and thrown out since negative logs are not allowed.

    About 7.7, your answer is correct.

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