August 18th, 2017, 06:26 PM  #1 
Newbie Joined: Aug 2017 From: Queensland Posts: 9 Thanks: 0  Not sure if index laws or log laws
Ok so I have another question  I'm struggling to get log laws and need to understand what's happening that I'm not grasping Express y in terms of x x=4(3^y) + 3^y  3^(y+log_3x) So my thought was to take 4 over to the left so it becomes x/4 = 3^y + 3^y  3(^y+log_3x) Then multiply out brackets x/4= 3^y+3^y3^y3^log_3x Then tidy up abit x/4=3^y3^log_3x Then to get powers down (this is where I'm confused) take log_3 of both sides log_3x/4=ylog_3(3)  log_3xlog_3(3) (and log_3(3) =1) so x/4=yx 2x/4=y x/4=y 
August 18th, 2017, 06:39 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,594 Thanks: 1492 
Apply $3^{a + b} = 3^a3^b$ first. Don't take logs until you need to. 
August 18th, 2017, 06:42 PM  #3 
Newbie Joined: Aug 2017 From: Queensland Posts: 9 Thanks: 0 
Now that I think about it (after writing it down) I think its more index laws rather than log laws (its just the power of log_3 has thrown me) , if I could get all to be a power of the same base (3) it would be easier wouldn't it?

August 18th, 2017, 06:59 PM  #4 
Senior Member Joined: May 2016 From: USA Posts: 904 Thanks: 359 
To get things into terms of y, you will eventually have to use logarithms, but first you must isolate terms involving y on one side of the equation and terms involving x on the other side of the equation. That is a basic technique for restating a relationship between x and y in terms in terms of x. $x = 4(3^y) + 3^y  3^{\{y + \log_3(x)\}} = 5 * 3^y  3^y * 3^{\log_3(x)} = 3^y\{5  \log_3(x)\}.$ $x \ne 243 \implies 5  \log_3(x) \ne 0 \implies 3^y = \dfrac{x}{5  \log_3(x)} \implies$ $\log_3 \left ( 3^y \right ) = \log_3 \left ( \dfrac{x}{5  \log_3(x)} \right ) \implies y \log_3(3) = \log_3 \left ( \dfrac{x}{5  \log_3(x)} \right ) \implies$ $y = \log_3 \left ( \dfrac{x}{5  \log_3(x)} \right ).$ Last edited by skipjack; August 18th, 2017 at 10:07 PM. Reason: to prefix "log" with "\" 
August 18th, 2017, 07:06 PM  #5 
Newbie Joined: Aug 2017 From: Queensland Posts: 9 Thanks: 0 
I don't get it 
August 18th, 2017, 07:13 PM  #6 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,518 Thanks: 506 Math Focus: Yet to find out.  
August 18th, 2017, 10:03 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,594 Thanks: 1492 
JeffM1 slipped up, so it's hardly surprising that PGibson doesn't understand it.

August 18th, 2017, 10:10 PM  #8 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,518 Thanks: 506 Math Focus: Yet to find out.  
August 19th, 2017, 12:11 PM  #9 
Senior Member Joined: May 2016 From: USA Posts: 904 Thanks: 359  No wonder. I goofed big time. $b,\ u > 0 \text { and } b \ne 1\ \text { and } v = b^{\log_b(u)} \implies$ $\log_b(v) = \log_b \left ( b^{\log_b(u)} \right ) = \log_b(u) * \log_b(b) \implies$ $\log_b(v) = \log_b(u) * 1 = \log_b(u) \implies v = u \implies$ $b^{\log_b(u)} = u.$ Now $x = 4(3^y) + 3^y  3^{\{y+\log_3(x)\}} = 3^y(5)  \left (3^y * 3^{\log_3(x)}\right ) =$ $3^y(5)  (3^y * x) = 3^y(5  x) \implies 3^y = \dfrac{x}{5  x} \text { if } x \ne 5.$ $0 < x < 5 \implies 0 < \dfrac{x}{5  x} = 3^y \implies$ $\log_3(3^y) = \log_3 \left ( \dfrac{x}{5  x} \right ) \implies y * \log_3(3) = \log_3 \left ( \dfrac{x}{5  x} \right ) \implies y = \log_3 \left ( \dfrac{x}{5  x} \right ).$ Last edited by skipjack; August 19th, 2017 at 08:25 PM. 
August 20th, 2017, 05:32 AM  #10  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,959 Thanks: 801  Quote:
Your first step should have been to use the fact that "4a+ a= 5a" to write the equation as x= 5(3^y) 3^(y+ log_3(x)). Then 3^(y+ log_3(x))= (3^y)(3^{log_3(x)}= (3^y)(x) because "a^x" and "log_a(x)" are inverse functions a^{log_a(x)}= x. But it is not just a matter of "log laws". You need to review basic algebra. Last edited by Country Boy; August 20th, 2017 at 05:37 AM.  

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