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    I have a few questions I'm stuck on and would appreciate some help.

    We have f(x)= ln(3x - 6) when x is any real no. and x>2

    How would i go about finding its range?

    f-1=(ex + 6)/3

    Since its range is f(x)'s domain, would the range be x>2? I have difficult in finding the range and domain sometimes. How would I find its domain?

    y=ln(3x-6), to calculate where it cuts the x-axis I'd plug in y=0 and solve accordingly, but what if I wanted to calcualte it by means of translation, like for example, we know that lnx cuts the x-axis at 1, so ln3x would cut it 1/3, hence ln(3x-6) would cut the x-axis at 6 1/3. But that answer doesn't seem to correspond with the other, where have I went wrong?

    And one more thing, how would I draw the graph of y=ln l3x - 6l.

    Rep will be dished out in due course!

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    (Original post by Dimez)
    I have a few questions I'm stuck on and would appreciate some help.

    We have f(x)= ln(3x - 6) when x is any real no. and x>2

    How would i go about finding its range?

    f-1=(ex + 6)/3

    Since its range is f(x)'s domain, would the range be x>2? I have difficult in finding the range and domain sometimes. How would I find its domain?

    y=ln(3x-6), to calculate where it cuts the x-axis I'd plug in y=0 and solve accordingly, but what if I wanted to calcualte it by means of translation, like for example, we know that lnx cuts the x-axis at 1, so ln3x would cut it 1/3, hence ln(3x-6) would cut the x-axis at 6 1/3. But that answer doesn't seem to correspond with the other, where have I went wrong?

    And one more thing, how would I draw the graph of y=ln l3x - 6l.

    Rep will be dished out in due course!

    Thanks
    domain of f is range of f^-1
    range of f is domain of f^-1
    if domain of f is x>2 then range of f^-1 is f^-1>2.
    this can be seen by
    f^-1=[e^x+6]/3=2+{e^x}/3
    since e^x/3>0 for all x clearly f^-1 is >2 for all values.
    the transformation ln(3x-6) depends on the order of the transformations.
    As you note if we do the stretch then translation it does not correspond.
    if we do translation then stretch we get
    ln(x-6)cuts at x= 7
    ln(3x-6) cuts at x=7/3
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    Thank you evariste.

    Can you help me with the latter part please, drawing the graph. I understand that its got something to do with relecting on the y-axis. Do I draw y=ln(3x -6) for x>0 then reflect on the y -axis? If that is it, the graph would emerge somewhat weird.

    Regards,


    Dimez
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    (Original post by Dimez)
    Thank you evariste.

    Can you help me with the latter part please, drawing the graph. I understand that its got something to do with relecting on the y-axis. Do I draw y=ln(3x -6) for x>0 then reflect on the y -axis? If that is it, the graph would emerge somewhat weird.

    Regards,


    Dimez
    The problem occurs at x<2, since 3x-6 is<0 and ln(3x-6) makes no sense.
    The mod signs rid us of that problem.
    x=2 will be assymptote.
    when x=7/3 y=0
    x=5/3 y=0
    start at top left of graph paper. curve downwards so it goes through 5/3,0 and goes down further as you get closer to x=2
    start at top right of graph paper, curve downwards so it goes through 7/3,0 and goes down further as you get closer to x=2.
 
 
 
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