Thanks MB!
I've stumbled across another question; Q) Solve .
I've reached , but the solutions show the answer as
I can see how it could work, but am a little confused as to the working out steps to get there. I'm pretty sure I'm just missing something very obvious.
Two ways you might get the answer using the base-2 logarithm:
Method 1: Rewrite the initial inequality in the following form:
Given what you've written about the answer you got, you should be able to proceed from there to the answer using the base-2 logarithm
Method 2: Convert your answer to the base-2 logarithm using the change of base formulae:
Then we have, and the result follows immediately:
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And another; Q)Solve
I got to and had no idea where to do to next. I checked the solutions and they went from there to;
and again I'm really confused how they just got rid of the natural log on both sides.
Then make both sides the exponent of
e. This is the inverse function of the natural logarithm, hence they "cancel out".
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And another! I'm really on a roll tonight; Q) Find inverse of function . I find it pretty easily; , but I'm wondering how you would identify that x = -3 asymtote just from looking at the inverse, when the log includes the translation as being divided by 5 inclusive of the dilation in the y-axis? If it even possible we'd be asked such a question in Methods? I'm just trying to overrprepare, and wondering how I'd identify it without having the pre-known knowledge that the asymtotes simply switch with inverses.
If you transform the plane by reflecting all points around the line y = x, then horizontal lines become vertical lines and vice versa. Hence, if the graph of an invertible function has a horizontal asymptote y = k, the graph of its inverse has a vertical asymptote x = h (and vice versa).
Alternatively, the vertical asymptote for a graph of y = ln(ax + b) can always be found by solving ax + b = 0.
Alternatively, if you work out the sequence of transformations mapping ln(x) to ln((x+3)5), you'll get x' = 5x – 3. You are correct to be suspicious about why the vertical asymptote remains at x = –3, because in general if you apply a dilation before a translation, you can't just find the asymptote of the image by applying
only the translation to the asymptote of the pre-image (consider, eg. applying x' = 5x – 3 to ln(x –1)). But for y = ln(x) the vertical asymptote
is the y-axis, so the dilation has no effect on points on this line; the only transformation that moves the asymptote is the translation.