+0

# Logs

+1
53
6
+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:

#1
+91049
0

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

Melody  Nov 9, 2017
#2
+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
+78753
+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

CPhill  Nov 9, 2017
edited by CPhill  Nov 9, 2017
edited by CPhill  Nov 10, 2017
#4
+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
+78753
0

Just reverse my steps......

CPhill  Nov 10, 2017
#6
+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?..