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January 24th, 2017, 02:38 PM   #1
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Log manipulation Question

If I had log 5 (x) = x - 3 and I wanted to move the x-3 all to the left hand side. What would the equation look like? Would it be (log 5 (x)) + 3 - x . This represents a horizontal translation correct?
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January 24th, 2017, 03:19 PM   #2
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$\log_5{x} = x-3 \implies \log_5{x} - (x-3) = 0$

no transformation occurs that I'm aware of other than subtracting the function $g(x)=x-3$ from the function $f(x) = \log_5{x}$
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January 24th, 2017, 03:35 PM   #3
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Ok thanks. And this transformation doesn't make sense to me.

If we have log 3 (x+9) + 2 = y. It states that we have a HT 9 units left and 2 units up. But if we manipulate that too

log 3 (x+9) = y - 2

And convert to Exp form. This gives me

3^(y-2) = x + 9 or 3^(y-2) - 9 = x

This looks to me more like a HT 2 units right and a vt 9 units down. I don't understand why x+9 is a HT when changing to exp form it looks like your K value (Vertical Translation) in exp form

Last edited by zekea; January 24th, 2017 at 03:41 PM.
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January 24th, 2017, 03:59 PM   #4
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the parent function for $y = \log_3(x+9) + 2$ is $y = \log_3{x}$

in succession (see first attached graph) ...

$\color{red}{y = \log_3(x+9)}$ shifts the parent function 9 units left

$y = \log_3(x+9) + 2$ shifts $y = \log_3(x+9)$ up 2 units

so, $\color{blue}{y = \log_3(x+9) + 2}$ shifts the parent function left 9 units and up 2
units


changing $y = \log_3(x+9) + 2$ to exponential form ...

$y - 2 = \log_3(x+9)$

$x+9 = 3^{y-2}$

$x = 3^{y-2} - 9$

the parent function is $x = 3^y$ which is shifted up 2 units and left 9 units because $x$ is a function of $y$

(see 2nd attached graph)

note the $\color{blue}{blue}$ graphs are identical
Attached Images
File Type: jpg log_trans.jpg (13.2 KB, 3 views)
File Type: jpg exp_trans.jpg (12.5 KB, 3 views)
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January 24th, 2017, 04:10 PM   #5
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The graph does show this to be true for sure, but looking at this in exp form confuses the heck out of me

$x = 3^{y-2} - 9$

because in my notes for exp form transformations the (y-2) should be your HT value (x-h) and the -9 should be K which would be a vertical translation according to the exp transformation description. Having a hard time relating this.

Like my notes have y = ac^b(x - h) + k where (x-h) is a horizontal translation and K is your vertical translation.

Last edited by zekea; January 24th, 2017 at 04:13 PM.
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January 24th, 2017, 04:27 PM   #6
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Quote:
Originally Posted by zekea View Post
The graph does show this to be true for sure, but looking at this in exp form confuses the heck out of me

$x = 3^{y-2} - 9$

because in my notes for exp form transformations the (y-2) should be your HT value (x-h) and the -9 should be K which would be a vertical translation according to the exp transformation description. Having a hard time relating this.

Like my notes have y = ac^b(x - h) + k where (x-h) is a horizontal translation and K is your vertical translation.
Your notes are correct if $y$ is a function of $x$, that is $y = f(x)$ ...

$f(x+c)$ is the function $f(x)$ shifted horizontally ... right if $c < 0$ and left if $c>0$

$f(x) + c$ is the function $f(x)$ shifted vertically ... down if $c < 0$ and up if $c>0$

now, understand that the transformations change if $x$ is a function of $y$, that is $x = f(y)$ ...

$f(y+c)$ is the function $f(y)$ shifted vertically ... up if $c < 0$ and down if $c>0$

$f(y) + c$ is the function $f(y)$ shifted horizontally ... left if $c < 0$ and right if $c>0$
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