May 20, 2013

Logarithm and exponential

  1. Solve the equation (1/2)2x + 1 = 1

  1. Solve x ym = y x3 for m.

  1. Given: log8(5) = b. Express log4(10) in terms of b.

  1. Simplify without calculator: log6(216) + [ log(42) – log(6) ] / log(49)

  1. Simplify without calculator: ((3-1 – 9-1) / 6)1/3

  1. Express (logxa)(logab) as a single logarithm.

  1. Find a so that the graph of y = logax passes through the point (e , 2).

  1. Find constant A such that log3 x = A log5x
    for all x > 0.

  1. Solve for x the equation log [ log (2 + log2(x + 1)) ] = 0

  1. Solve for x the equation 2 x b4 logbx = 486

  1. Solve for x the equation ln (x – 1) + ln (2x – 1) = 2 ln (x + 1)

  1. Find the x intercept of the graph of y = 2 log( sqrt(x – 1) – 2)

  1. Solve for x the equation 9x – 3x – 8 = 0

  1. Solve for x the equation 4x – 2 = 3x + 4

15. If logx(1 / 8) = -3 / 4, than what is x?

 

Solutions to the Above Problems

  • Rewrite equation as (1/2)2x + 1 = (1/2)0
    Leads to 2x + 1 = 0
    Solve for x: x = -1/2


  • Divide all terms by x y and rewrite equation as: ym – 1 = x2
    Take ln of both sides (m – 1) ln y = 2 ln x
    Solve for m: m = 1 + 2 ln(x) / ln(y)


  • Use log rule of product: log4(10) = log4(2) + log4(5)
    log4(2) = log4(41/2) = 1/2
    Use change of base formula to write: log4(5) = log8(5) / log8(4) = b / (2/3) , since log8(4) = 2/3
    log4(10) = log4(2) + log4(5) = (1 + 3b) / 2


  • log6(216) + [ log(42) – log(6) ] / log(49)
    = log6(63) + log(42/6) / log(72)
    = 3 + log(7) /2 log(7) = 3 + 1/2 = 7/2


  • ((3-1 – 9-1) / 6)1/3
    = ((1/3 – 1/9) / 6)1/3
    = ((6 / 27) / 6)1/3 = 1/3


  • Use change of base formula: (logxa)(logab)
    = logxa (logxb / logxa) = logxb


  • 2 = logae
    a2 = e
    ln(a2) = ln e
    2 ln a = 1
    a = e1/2


  • Use change of base formula using ln to rewrite the given equation as follows
    ln (x) / ln(3) = A ln(x) / ln(5)
    A = ln(5) / ln(3)


  • Rewrite given equation as: log [ log (2 + log2(x + 1)) ] = log (1) , since log(1) = 0.
    log (2 + log2(x + 1)) = 1
    2 + log2(x + 1) = 10
    log2(x + 1) = 8
    x + 1 = 28
    x = 28 – 1


  • Note that b4 logbx = x4
    The given equation may be written as: 2x x4 = 486
    x = 2431/5 = 3


  • Group terms and use power rule: ln (x – 1)(2x – 1) = ln (x + 1)2
    ln function is a one to one function, hence: (x – 1)(2x – 1) = (x + 1)2
    Solve the above quadratic function: x = 0 and x = 5
    Only x = 5 is a valid solution to the equation given above since x = 0 is not in the domain of the expressions making the equations.


  • Solve: 0 = 2 log( sqrt(x – 1) – 2)
    Divide both sides by 2: log( sqrt(x – 1) – 2) = 0
    Use the fact that log(1)= 0: sqrt(x – 1) – 2 = 1
    Rewrite as: sqrt(x – 1) = 3
    Raise both sides to the power 2: (x – 1) = 32
    x – 1 = 9
    x = 10


     

  • Given: 9x – 3x – 8 = 0
    Note that: 9x = (3x)2
    Equation may be written as: (3x)2 – 3x – 8 = 0
    Let y = 3x and rewite equation with y: y2 – y – 8 = 0
    Solve for y: y = ( 1 + sqrt(33) ) / 2 and ( 1 – sqrt(33) ) / 2
    Since y = 3x, the only acceptable solution is y = ( 1 + sqrt(33) ) / 2
    3x = ( 1 + sqrt(33) ) / 2
    Use ln on both sides: ln 3x = ln [ ( 1 + sqrt(33) ) / 2]
    Simplify and solve: x = ln [ ( 1 + sqrt(33) ) / 2] / ln 3


     

  • Given: 4x – 2 = 3x + 4
    Take ln of both sides: ln ( 4x – 2 ) = ln ( 3x + 4 )
    Simplify: (x – 2) ln 4 = (x + 4) ln 3
    Expand: x ln 4 – 2 ln 4 = x ln 3 + 4 ln 3
    Group like terms: x ln 4 – x ln 3 = 4 ln 3 + 2 ln 4
    Solve for x: x = ( 4 ln 3 + 2 ln 4 ) / (ln 4 – ln 3) = ln (34 * 42) / ln (4/3) = ln (34 * 24) / ln (4/3)
    = 4 ln(6) / ln(4/3)


  • Rewrite the given equation using exponential form: x– 3 / 4 = 1 / 8
    Raise both sides of the above equation to the power -4 / 3: (x– 3 / 4)– 4 / 3 = (1 / 8)– 4 / 3
    simplify: x = 84 / 3 = 24 = 16