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Math Help - bunch of Logarithm qusetions help^^

  1. #1
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    bunch of Logarithm qusetions help^^

    If log3 4=X express each of the following in terms of x

    a) log3 16-2log3 32 b)log3 2

    Express y as a function of x. state the domain
    a)logY = log(1-3x)
    b)log2 (y-3) = 5 + log2 (x-4)

    solve and check
    a)2logM + 3logM =10

    Hi i am totally stuck with those log questions
    can anyone show me how to do it in full steps?

    i tried the last one 2logM + 3logM =10 i got stuck after moving the 2 to the power of M.

    for the other ones i dont have any idea on how to start on it..ty
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  2. #2
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    Re :

    (1) Given log_3 4=x

    (a) log_3 16-2log_3 32
    = log_3 4^2-2log_3 2^5
    = log_3 4^2-log_3 2^{10}
    =log_3 4^2-log_3 (4^\frac{1}{2})^{10}
    Can you continue from here ?

    (b) log_3 2=\frac{1}{2}log_3 4 = 1/2x

    (2) lg y= lg(1-3x)
    Cancel logs from both sides , then y=1-3x

    log_2 (y-3)=5+log_2(x-4)
    log_2 (y-3)=log_2 2^5+log_2(x-4)
    log_2 (y-3)=log_2(2^5(x-4))
    Take away logs from both sides .....

    (3) 2lg M +3lg M =10
    lgM^2+lgM^3=10
    lgM^5=10
    Try continue from here ...
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  3. #3
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    1)

    a)u cancel the logs than turn 4^(1/2)^10 into 4^5?
    then 4^(2-5) like this? Then i got stuck again >.< sry

    and for B how did u get 1/2x right away?

    2)i dont understand what "y as a function of x" means. What are you actually looking for?

    3)
    <br /> <br />
    " alt="lgM^5=10
    " />

    5=M^10
    then i got stuck again>.< do you square root the 5 10 times?

    >.<ty
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  4. #4
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    Quote Originally Posted by hovermet View Post
    1)

    a)u cancel the logs than turn 4^(1/2)^10 into 4^5?
    then 4^(2-5) like this? Then i got stuck again >.< sry

    and for B how did u get 1/2x right away?

    2)i dont understand what "y as a function of x" means. What are you actually looking for?

    3)
    <br /> <br />
    " alt="lgM^5=10
    " />

    5=M^10
    then i got stuck again>.< do you square root the 5 10 times?

    >.<ty
    1. Keep in mind that \boxed{x = \log_3(4)}

    2. From

    <br />
=log_3 4^2-log_3 (4^\frac{1}{2})^{10} = \log_3(4^2)-\log_3{4^5} = 2\log_3(4)-5\log_3(4) = 2x-5x = -3x<br />

    3. I guess 1)(b) is obvious right now. (See my initial remark)

    4. You are asked to derive an equation where y is expressed by x. In other words: Solve the given equation for y and look what happens to the x.

    a) \log Y = \log(1-3x) The base of these logs isn't specified so you can use any base. (Mathaddict has used the base 10). Use the logs as exponents to a base and you'll get:

    y = 1-3x This equation describes a function y of x.

    b) \log_2 (y-3) = 5 + \log_2 (x-4) The base is 2. Use this base and the given logarithms (=exponents) to get rid of the logarithms:

    2^{\log_2 (y-3)}=2^{5 + \log_2 (x-4)} ~\implies~y-3 = 2^5 \cdot (x-4)~ \implies~y-3=32x-128~\implies~\boxed{y=32x-125}

    solve and check
    a) 2 \log M + 3 \log M =10~\implies~5\log M =10~\implies~\log M = 2

    If you use \log (n)= \log_{10}(n) then the equation becomes: \boxed{M = 100}
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  5. #5
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    ty>.<

    but i still don't understand the first question

    how did u get 2\log_3(4)-5\log_3(4) = 2x-5x = -3x? where did the X came from and can't u just add the exponents? since the have the same base. also on my text the answer is suppose to be 1 + 3X why?

    and i got another question

    log3 2 = x (same as question one

    a)log3 8


    i got the the point where log8/log3 =x after i done that i got 1.8xxxxx. but the answer is 3X..
    ============

    log3 4=x

    a)log3 16- 2log3 32
    i got to the point where 3log3 4 - 8log3 4 and came out with an answer -5X, but in the text it saids the answer is -3X....help>.<
    ================
    slove for X
    log8 x-1 = log8 (x-1)

    i canceled log8 out from both sides and got x-1 = x-1...... ?= -2 stuck...why can't u just cancel both sides and find the answer>.< log is so complicated>.< help
    ====================
    solve and check

    log4 (x+2) + log4 ( x-1) =1
    got stuck when i got to x^2+ x-3=0 can't factor..

    Last edited by hovermet; January 7th 2009 at 08:54 PM.
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  6. #6
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    Quote Originally Posted by hovermet View Post
    If log3 4=X express each of the following in terms of x

    ...
    Quote Originally Posted by hovermet View Post
    ty>.<

    but i still don't understand the first question

    how did u get 2\log_3(4)-5\log_3(4) = 2x-5x = -3x? where did the X came from and can't u just add the exponents? since the have the same base. also on my text the answer is suppose to be 1 + 3X why?

    ...
    You have given this term in your first post
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  7. #7
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    Quote Originally Posted by hovermet View Post
    ...
    log3 2 = x (same as question one

    a)log3 8


    i got the the point where log8/log3 =x after i done that i got 1.8xxxxx. but the answer is 3X..
    ============

    ...
    If \log_3(2) = x then

    \log_3(8)=\log_3(2^3)=3\log_3(2) = 3x

    (and don't ask me where the x comes from ... !)
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  8. #8
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    Quote Originally Posted by hovermet View Post
    ...
    log3 4=x

    a)log3 16- 2log3 32
    i got to the point where 3log3 4 - 8log3 4 and came out with an answer -5X, but in the text it saids the answer is -3X....help>.< I answered this question already. See post #4 of this thread.
    ================
    slove for X
    log8 x-1 = log8 (x-1)
    i canceled log8 out from both sides and got x-1 = x-1...... ?= -2 stuck...why can't u just cancel both sides and find the answer>.< log is so complicated>.< help
    ====================
    \log_8(x) - 1 = \log_8(x-1)~\implies~x\cdot \frac18=x-1~\implies~-\dfrac78 x=- 1~\implies~x= \dfrac87

    Quote Originally Posted by hovermet View Post
    solve and check

    log4 (x+2) + log4 ( x-1) =1
    got stuck when i got to x^2+ x-3=0 can't factor..
    \log_4 (x+2) + \log_4 ( x-1) =1~\implies~(x+2)(x-1)=4~ \implies~x^2+x-6=0~\implies~x=-3~\vee~x=2

    Since the domain of this equation is d=(1, \infty) x = -3 isn't a solution of the equation.
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