Help on these problems

Discussion in 'Calculus' started by Scoota, Jul 3, 2023.

  1. Scoota

    Scoota

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    I tried but couldn't solve
     

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    Scoota, Jul 3, 2023
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  2. Scoota

    HallsofIvy

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    To solve [math]log_4^2(x)+ log_4(\sqrt{x})= 15[/math] first note that [math]log_4(\sqrt{x})= k=log_4(x^{1/2})= (1/2) log_4(x)[/math] so [math]log_4^2(x)+ (1/2)log_4(x)= 15[/math].
    Now let [math]y= log_4(x)[/math] so [math]y^2+ (1/2)y= 15[/math].
    [math]2y^2+ y- 15= (2y- 5)(y+ 3)= 0[/math]. y= 5/2 or y= -3.

    A logarithm is never negative so [math]log_4(x)= 5/2[/math] and [math]x= 4^{5/2}= 2^5= 32.

    Any logarithm is defined only for positive numbers so the domain of [math]\log_3^2(x)+ 2 log_3(x)= 3[/math] is x> 0. To solve it, let [math]y= log_3(x)[/math] so [math]y^2+ 2y= 3[/math]. [math]y^2+ 2y- 3= (y+ 3)(y- 1)= 0[/math]. y= -3 or y= 1.

    Again, a logarithm is never negative so y= log_3(x)= 1 and x= 3.
     
    HallsofIvy, Aug 4, 2023
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