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Logs and Exponentials?

I’m unsure on how I would get two possible answers for this?? Please help

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3 Answers

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  • 1 month ago
    Favorite Answer

    1 + 2 log₃(x) = log₃(28x - 9)

    We can move the log on the left side to the right by subtracting that term from both sides:

    1 = log₃(28x - 9) - 2 log₃(x)

    Next, I'll move the "2" that's outside the second log inside by putting it as an exponent to what's inside:

    1 = log₃(28x - 9) - log₃(x²)

    The difference of two logs with the same base is the same as the log of the quotient, so:

    1 = log₃[(28x - 9) / x²]

    Now if we make both sides of the equation an exponent over a base of 3 this will cancel out the log and leave only what's inside the log:

    3¹ = (28x - 9) / x²

    3 = (28x - 9) / x²

    Multiply both sides by x² then we can simplify this as a quadratic:

    3x² = 28x - 9

    3x² - 28x + 9 = 0

    I'll use the quadratic equation:

    x = [ -b ± √(b² - 4ac)] / (2a)

    x = [ -(-28) ± √((-28)² - 4(3)(9))] / (2 * 3)

    x = [ 28 ± √(784 - 108)] / 6

    x = [ 28 ± √(676)] / 6

    x = (28 ± 26) / 6

    x = 2/6 and 54/6

    x = 1/3 and 9

    Testing the solutions to make sure they are correct:

    1 + 2 log₃(x) = log₃(28x - 9)

    1 + 2 log₃(1/3) = log₃(28 * 1/3 - 9) and 1 + 2 log₃(9) = log₃(28 * 9 - 9)

    1 + 2 log₃(1/3) = log₃(28/3 - 9) and 1 + 2 log₃(9) = log₃(252 - 9)

    1 + 2 log₃(1/3) = log₃(28/3 - 27/3) and 1 + 2 log₃(9) = log₃(243)

    1 + 2 log₃(1/3) = log₃(1/3) and 1 + 2 log₃(9) = log₃(243)

    1/3, 9, and 243 are all powers of 3:

    1 + 2 log₃(3⁻¹) = log₃(3⁻¹) and 1 + 2 log₃(3²) = log₃(3⁵)

    1 + 2(-1) = -1 and 1 + 2(2) = 5

    1 - 2 = -1 and 1 + 4 = 5

    -1 = -1 and 5 = 5

    TRUE and TRUE

    Both are solutions, so your answer is D from your multiple choice.

  • 1 month ago

    1 + 2log₃x = log₃(28x – 9)

    log₃3 = 1

    log₃3 + 2log₃x = log₃(28x – 9)

    log₃3 + log₃x² = log₃(28x – 9)

    log₃(3x²) = log₃(28x – 9)

    3x² = 28x – 9

    3x² – 28x + 9 = 0

    quadratic equation:

    to solve ax² + bx + c = 0

    x = [–b ± √(b²–4ac)] / 2a

    x = [28 ± √(28²–4•3•9)] / 6

    x = [28 ± √(784–108)] / 6

    x = [28 ± √(676)] / 6

    x = [28 ± 26] / 6

    x = 1/3, 9

  • Alan
    Lv 7
    1 month ago

    1  =log_b3(3) 

    log_b3(3) + 2*log_b3(x)   = log_b3(28x-9)

    log_b3(3)  + log_b3(x^2)  =log_b3(28x-9) 

    log_b3(  3 *x^2)     =  log_b3( 28x-9)   

    therefore 

    3x^2  = 28x -9

    3x^2 -28x + 9 = 0  

    factor or use quadratic formula 

    (3x -a)(x - b)     

    (3x - 1) (x -9)  =0     

    so either 

    3x -1 = 0 

    or 

    x -9 = 0 

    3x = 1  

    x= 1/3 

    or 

    x = 9 

    so D is correct 

     

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