+0  
 
+1
52
6
avatar+134 

Hi good people!,

 

this log problem is confusing me:

 

It says: write in expanded log form:

 

\(log{a^4 \over3^2x}\)

 

So, I do this:

 

\(loga^4-(log3^2+logx)\)

 

but this gives \(loga^4-log3^2-logx \)

which gives: \(4loga-2log3-logx\)

 

If I convert the last line back to a solve, I cannot see how I will get  \(log{a^4 \over3^2x}\), again?..

Where am I going wrong?

juriemagic  Nov 9, 2017
Sort: 

6+0 Answers

 #1
avatar+91047 
0

Looks right to me, I do not understand what your question is. 

Melody  Nov 9, 2017
 #2
avatar+134 
+1

Hi Melody,

 

well, it's this:

 

\(4loga-2log3-logx\)

 goes to:

 

\(log{a^4 \over 3^2}\)

 

the "-logx"...how do i "see" mathematically that it is supposed to join the \(3^2\) ?

juriemagic  Nov 9, 2017
edited by juriemagic  Nov 9, 2017
 #3
avatar+78744 
+1

Let's reverse the process

 

log  ( a4 / [ 32 x ]  )     =

 

log a4  -  [ log (32 * x) ] =

 

log a 4 - [ log 32 + log x ]   =

 

4 log a  - [ 2 log 3  + log x ]

 

4 log a  - 2 log 3  - log x

 

 

cool cool cool

CPhill  Nov 9, 2017
edited by CPhill  Nov 9, 2017
edited by CPhill  Nov 10, 2017
 #4
avatar+134 
0

Hi CPhil,

 

yes, that is the part I understand, what I do not get is how to reverse that back to the single log term?

juriemagic  Nov 10, 2017
 #5
avatar+78744 
0

Just reverse my steps......smiley smiley smiley

 

 

cool cool cool

CPhill  Nov 10, 2017
 #6
avatar+134 
+1

I see,

 

so it's a rule to take the second and third terms, should there be a "-" between them, together into a bracket, which obviously will have the "-" change to a "+"...and move on from there?..

 

Thanx for your time CPhill...

juriemagic  Nov 10, 2017

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