My Math Forum Not sure if index laws or log laws

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 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^y-3^y-3^log_3x Then tidy up abit x/4=3^y-3^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=y-x 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. Thanks from PGibson
 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. $x = 4(3^y) + 3^y - 3^{\{y + \log_3(x)\}} = 5 * 3^y - 3^y * 3^{\log_3(x)} = 3^y\{5 - \log_3(x)\}.$ $x \ne 243 \implies 5 - \log_3(x) \ne 0 \implies 3^y = \dfrac{x}{5 - \log_3(x)} \implies$ $\log_3 \left ( 3^y \right ) = \log_3 \left ( \dfrac{x}{5 - \log_3(x)} \right ) \implies y \log_3(3) = \log_3 \left ( \dfrac{x}{5 - \log_3(x)} \right ) \implies$ $y = \log_3 \left ( \dfrac{x}{5 - \log_3(x)} \right ).$ Thanks from PGibson 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
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Quote:
 Originally Posted by PGibson I don't get it
Which part? Try not to just gloss over all the 'math' as it can be daunting sometimes. Slowly work through each line, and try identify the part that you don't understand.

 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
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 Originally Posted by skipjack JeffM1 slipped up, so it's hardly surprising that PGibson doesn't understand it.
Right. I didn't actually read it myself (should take some of my own advice!). In any case, my point still stands.

August 19th, 2017, 12:11 PM   #9
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 Originally Posted by PGibson I don't get it
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
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Quote:
 Originally Posted by PGibson 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)
Your very first step is wrong and has nothing to do with "logarithm laws" or "exponential laws"! You cannot "take 4 over to the left" because it is multiplying only the first "(3^y)" not the entire expression.

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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