anonymous
  • anonymous
Simplify 5 log2 k − 8 log2 m + 10 log2 n. 7 log2 (k − m + n) 7 log2 kn over m log2 50 kn over 8 m log2 k to the fifth power n to the tenth power over m to the eighth power
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@uri @LeAnn01Xxx @e.mccormick @Lady.Liv1776 @TylerMJ325 @Terrorclaw @Phebe @keshaun455 @JoshKoikkara
anonymous
  • anonymous
Hmm... 5*log2 k - 8*log2 m + 10*log2 n log2 (k^5*n^10)/m^8 , I'm not sure...do u get the same?
anonymous
  • anonymous
I don't want the answer I want to learn how to do it. I'm really bad at Maths ...

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Nnesha
  • Nnesha
quotient rule\[\large\rm log_b y - \log_b x = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication power rule \[\large\rm log_b x^y = y \log_b x\] remember!
anonymous
  • anonymous
@Marcos_RG4L Hmm, I forgot Latex wait.... http://openstudy.com/study#/updates/4ffb5ce1e4b00c7a70c465e9
anonymous
  • anonymous
@Nnesha Hmm...Great !!
Nnesha
  • Nnesha
first apply power property \[\large\rm log_b x^y = y \log_b x\]
anonymous
  • anonymous
5 log 2 ?
Nnesha
  • Nnesha
|dw:1434389999658:dw| 2 is a base number at front of log became exponent of k
Nnesha
  • Nnesha
\[\huge\rm 5 \log_{2} = k = \log_2 k^5\]
anonymous
  • anonymous
Ok Always change ?
Nnesha
  • Nnesha
\[\huge\rm 5 \log_{2} k = \log_2 k^5\]
Nnesha
  • Nnesha
yes always so what is \[8 \log_2 m = ???\]
Nnesha
  • Nnesha
|dw:1434390120641:dw|
anonymous
  • anonymous
ok
Nnesha
  • Nnesha
\[\huge\rm 8 \log_2 m = ???\]try this one
anonymous
  • anonymous
m= 2 log 8^k?
Nnesha
  • Nnesha
there isn't any k in that part
anonymous
  • anonymous
Then is only the 8 log 2^ m = 2 log 8^ ?
Nnesha
  • Nnesha
look at the log property \[\large\rm y \log_b x = log_b x^y\] number at front of log became exponent of x no the base
Nnesha
  • Nnesha
here is an example \[\huge\rm \color{red}{3} \log_4 \color{blue}{ 7}\]\[\large\rm log_4 \color{blue}{7}^\color{reD}{3}\]
Nnesha
  • Nnesha
\[\huge\rm \color{reD}{8} \log_2 \color{blue}{m} = ???\]
anonymous
  • anonymous
8 log 2 m = log 2 m 8^ ?
Nnesha
  • Nnesha
yay!!
anonymous
  • anonymous
Ok, The colors help me ...
Nnesha
  • Nnesha
\[\huge\rm log_2 k^5 - \log_2 m^8 + \color{red}{10} \log_2 \color{blue}{n}\] last one apply power rule
anonymous
  • anonymous
10 log 2 n = log 2 n 10^ ?
Nnesha
  • Nnesha
yes right \[\huge\rm log_2 k^5 \color{ReD}{-} \log_2 m^8 \color{blue}{+} \log_2 {n^{10}}\] now look at the addition and subtraction sign sign and look at the properties which one you should apply ?
anonymous
  • anonymous
Multiplication?
Nnesha
  • Nnesha
for which one there are two part negative and plus sign
anonymous
  • anonymous
Idk umm. Log 2 m 8^ x log 2 n 10^?
Nnesha
  • Nnesha
log_2 m^8 is negative so which property is for negative signs ?
Nnesha
  • Nnesha
you can write it as \[\huge\rm log_2 k^5 + \log_2 n^{10} - \log_2 m^8\]
anonymous
  • anonymous
ok
Nnesha
  • Nnesha
you can write it as \[\huge\rm \color{reD}{log_2 k^5 + \log_2 n^{10}} \color{blue}{ - \log_2 m^8}\] for red part = apply product property for blue part = apply quotient property
anonymous
  • anonymous
Thanks for ur help I need to go thank you very much...
Nnesha
  • Nnesha
alright gO_Od luck!

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