SOLUTION: {{{ -log( 3,x ) +2 }}}
I have figured out that the +2 is not (x+2) but -log(3x)+2. I know that the x-intercept for this logarithm is (9,0) and that this log graphs in Quadrant I
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-> SOLUTION: {{{ -log( 3,x ) +2 }}}
I have figured out that the +2 is not (x+2) but -log(3x)+2. I know that the x-intercept for this logarithm is (9,0) and that this log graphs in Quadrant I
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Question 446668:
I have figured out that the +2 is not (x+2) but -log(3x)+2. I know that the x-intercept for this logarithm is (9,0) and that this log graphs in Quadrant I with a vertical asymptote being the y-axis. The graph crosses into quadrant IV through point (9,0).
I am suppose to convert and plot this logarithm as an exponent. Then using the X,Y points I devise, invert and use to plot the logarithm.
I am not able to convert to an exponent as I do not know how to handle the +2. I can't seem to reverse engineer this using the points (9,0) either as the negative sign prevents this.
I'm pretty stuck. I would really love some kind of hints to solving this one.
Thank you so much for your time and expertise!! Found 2 solutions by stanbon, scott8148:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! y = -log(3x)+2
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Determine the inverse:
1st: Interchange x and y.
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x = -log(3y)+2
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2nd: Solve for "y":
-log(3y) = x-2
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log(3y) = 2-x
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3y = 10^(2-x)
y = (1/3)*10^(2-x)
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That is the inverse function.
3rd: Graph both on the same set of axes:
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Cheers,
Stan H.
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