katherinesmith
  • katherinesmith
Write 1 – 3log5x as a single logarithm.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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katherinesmith
  • katherinesmith
\[1 - 3\log _{5}x\]
anonymous
  • anonymous
what is 1 in log5 form?
katherinesmith
  • katherinesmith
i'm not sure

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More answers

Luigi0210
  • Luigi0210
Here, think of it this way: \[\log_{5}x=1\] \[5^1=x\]
katherinesmith
  • katherinesmith
so 5?
anonymous
  • anonymous
yes
anonymous
  • anonymous
now add \[\log_{5}5 - 3\log_{5}x\]
anonymous
  • anonymous
subtract sorry
katherinesmith
  • katherinesmith
a little help please
anonymous
  • anonymous
help please @Luigi0210
Luigi0210
  • Luigi0210
Okay, now change that second log, do remember this rule: \[{\huge ylog_{k}x=\log_{k}x^y }\]
katherinesmith
  • katherinesmith
okay...
anonymous
  • anonymous
\[\log_{5} 5 -\log_{5} x ^{3}\]
anonymous
  • anonymous
subtraction is division, right? |dw:1377732866434:dw|
anonymous
  • anonymous
I am not sure.
katherinesmith
  • katherinesmith
i really have no idea what to do.
katherinesmith
  • katherinesmith
@zepdrix
Luigi0210
  • Luigi0210
That's right
zepdrix
  • zepdrix
Confused by the logs girl? D:
katherinesmith
  • katherinesmith
yes :(
katherinesmith
  • katherinesmith
start me off from the beginning and work through it because there's no sense of pattern here and i'm lost
zepdrix
  • zepdrix
|dw:1377733060518:dw|5 to what power gives us 5? D:
katherinesmith
  • katherinesmith
so how do i write that as a "single logarithm?" can i write it like 5^? = 5 and that be my answer?
zepdrix
  • zepdrix
No. I was just trying to give you some understanding of how logs work. :(
katherinesmith
  • katherinesmith
ugh...
zepdrix
  • zepdrix
Ok here is another way we can approach it.
katherinesmith
  • katherinesmith
i don't know what they're asking for. i don't know what a single logarithm looks like.
zepdrix
  • zepdrix
Just try to remember this little fact. If our `base` ever matches the `contents` of our log, that is equivalent to 1. Examples:\[\Large \log_{10}10=1\]\[\Large \log_33=1\]
zepdrix
  • zepdrix
\[\Large \color{royalblue}{1}-3\log_5x\]We want to apply this idea `in reverse`. \[\Large \color{royalblue}{1=\log_55}\]Plugging this in gives us,\[\Large \color{royalblue}{\log_55}-3\log_5x\]
katherinesmith
  • katherinesmith
okay i understand that.
zepdrix
  • zepdrix
Once we've taken care of that tricky step all we need to do is apply a couple of `Rules of Logarithms` to simply.
zepdrix
  • zepdrix
First rule we'll apply:\[\Large \color{#CC0033}{b\cdot\log(a)=\log(a^b)}\]
zepdrix
  • zepdrix
Undestand how that will change our second term?
katherinesmith
  • katherinesmith
can you plug in the numbers so i know which number corresponds to which letter
zepdrix
  • zepdrix
Applying this rule to the second term:\[\Large 3\cdot\log_5(x) \qquad=\qquad \log_5(x^3)\]
zepdrix
  • zepdrix
Ignore the base of the log, it has nothing to do with this rule. It involves the 3 and the x.
katherinesmith
  • katherinesmith
okay
zepdrix
  • zepdrix
So that brings us to here:\[\large \log_55-3\log_5x \quad=\quad \log_55-\log_5(x^3)\]
zepdrix
  • zepdrix
The other rule of logarithms that we want to apply is:\[\Large \color{#CC0033}{\log(a)-\log(b) \quad=\quad \log\left(\frac{a}{b}\right)}\]
katherinesmith
  • katherinesmith
alright
zepdrix
  • zepdrix
Soooo what does that give you? :3
katherinesmith
  • katherinesmith
i have no idea.
zepdrix
  • zepdrix
:c
katherinesmith
  • katherinesmith
like what.

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