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Question on logs???

log base 2 of 56 minus log base 4 of 49. i know the answer is 3 but i need to show my work. please help!

3 Answers

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  • 1 decade ago
    Favorite Answer

    To perform operations on logs, they must have the same base. Because 4=2², log(4)49=½log(2)49

    =log(2)49^(½)

    =log(2)sqrt(49)

    =log(2)7

    Now, log(2)56-log(2)7 = log(2)56/7

    =log(2)8=3

  • Puggy
    Lv 7
    1 decade ago

    Let x = log[base 2](56), and

    Let y = log[base 4](49)

    We can use the change of base formula on log[base 4](49). Remember that

    log[base a](c) = log[base b](c) / log[base b](a)

    Let's change this to base 2.

    log[base 4](49) = log[base 2](49) / log[base 2](4)

    But log[base 2](4) = 2, so we have

    log[base 4](49) = (1/2) log[base 2](49)

    We can move the (1/2) as a power of 49 as per the log property

    c log[base b](a) = log[base b](a^c).

    log[base 4](49) = log[base 2](49^(1/2))

    49^(1/2) is equal to 7, so we have

    log[base 2](7)

    Remember that originally we have

    log[base 2](56) - log[base 4](49)

    and we just discovered that log[base 4](49) = log[base 2](7), so

    log[base 2](56) - log[base 2](7)

    Now we can combine these two logs as per the log property

    log[base 2](a) - log[base 2](c) = log[base 2](a/c), giving us

    log[base 2](56/7)

    log[base 2](8)

    This is asking, "2 to the what power is equal to 8?" to which the answer is 3.

  • Anonymous
    1 decade ago

    u can solve it -

    log 56 - (log 49)/(log 4) 1. ALL base 2

    becauz log c (base a) = log c (base b)/log b (base a)

    so, 1. =

    log 56 - (log 49)/2

    = (2 log 56 - log 49)/2

    = (log (56^2) - log 49)/2

    = log(56^2 / 49)/2

    = (log 64) /2

    = 6/2

    =3

    all done by log rules, u can search for them

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