January 24th, 2017, 02:38 PM  #1 
Newbie Joined: Dec 2016 From: Canada Posts: 12 Thanks: 0  Log manipulation Question
If I had log 5 (x) = x  3 and I wanted to move the x3 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?

January 24th, 2017, 03:19 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,773 Thanks: 1427 
$\log_5{x} = x3 \implies \log_5{x}  (x3) = 0$ no transformation occurs that I'm aware of other than subtracting the function $g(x)=x3$ from the function $f(x) = \log_5{x}$ 
January 24th, 2017, 03:35 PM  #3 
Newbie Joined: Dec 2016 From: Canada Posts: 12 Thanks: 0 
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^(y2) = x + 9 or 3^(y2)  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. 
January 24th, 2017, 03:59 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,773 Thanks: 1427 
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 ... $y  2 = \log_3(x+9)$ $x+9 = 3^{y2}$ $x = 3^{y2}  9$ 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 
January 24th, 2017, 04:10 PM  #5 
Newbie Joined: Dec 2016 From: Canada Posts: 12 Thanks: 0 
The graph does show this to be true for sure, but looking at this in exp form confuses the heck out of me $x = 3^{y2}  9$ because in my notes for exp form transformations the (y2) should be your HT value (xh) 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 (xh) is a horizontal translation and K is your vertical translation. Last edited by zekea; January 24th, 2017 at 04:13 PM. 
January 24th, 2017, 04:27 PM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,773 Thanks: 1427  Quote:
$f(x+c)$ 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 $c>0$  

Tags 
log, manipulation, question 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How is this algebraic manipulation possible?  shreddinglicks  Algebra  5  October 10th, 2014 03:59 AM 
Sigma manipulation  Keroro  Algebra  4  June 10th, 2012 06:18 AM 
Algebraic Manipulation  tinku  Algebra  3  February 23rd, 2012 06:57 PM 
Integral Manipulation.  ZardoZ  Real Analysis  2  May 27th, 2011 06:20 PM 
Algebraic Manipulation...  Liqwde  Algebra  4  September 1st, 2010 03:53 PM 