SPHott51
  • SPHott51
Use the properties of logarithms to evaluate
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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SPHott51
  • SPHott51
|dw:1447126622180:dw|
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \log_26+\log_232-\log_224 }\) this is your eq, right?
SPHott51
  • SPHott51
Yes

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \log_ab+\log_ac=\log_a(b\cdot c) }\) \(\large\color{black}{ \displaystyle \log_ab-\log_ac=\log_a(b\div c) }\)
SolomonZelman
  • SolomonZelman
knowing these two properties, you get: \(\large\color{black}{ \displaystyle \log_2(6\times 32\div 24) }\)
SolomonZelman
  • SolomonZelman
And then, after you simplify that, another good thing to know is: \(\large\color{black}{ \displaystyle \log_a(a^b) =b\log_a(a)=b\times 1 = b }\)
SPHott51
  • SPHott51
So log2 192 / log2 24 ?
SolomonZelman
  • SolomonZelman
Ok, lets start again, step by step.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \log_26+\log_232-\log_224 }\) RULE OF ADDITION: \(\large\color{blue}{ \displaystyle \log_ab+\log_ac= \log_a(b\times c)}\) So, apply this rule to the first two terms and you get: \(\large\color{black}{ \displaystyle \log_2(6\times 32)-\log_224 }\) NEXT, RULE OF SUBTRACTION: \(\large\color{blue}{ \displaystyle \log_ab-\log_ac= \log_a(b\div c)}\) So, apply this rule and you get: \(\large\color{black}{ \displaystyle \log_2\left[(6\times 32)\div24\right] }\)
SPHott51
  • SPHott51
So then I would use my graphic calculater ? As in ((log 6 *log32)/log24)) ?
SolomonZelman
  • SolomonZelman
6 × 32 ÷ 24 = (3 × 2 × 4 × 8) ÷ (3 × 2 × 4) = 8. So you then get: \(\large\color{black}{ \displaystyle \log_2\left[8\right] }\)
SolomonZelman
  • SolomonZelman
And then: \(\large\color{black}{ \displaystyle \log_2\left[8\right]=\log_2\left[2^3\right]=3\times \log_2\left[2\right]=\quad ? }\)
SPHott51
  • SPHott51
Yeah I got 8 too but it says its wrong.... Okay so.. 3 * log2 (2) ... Uh...
SolomonZelman
  • SolomonZelman
\(\color{brown}{\log_a(a)=1}\) And thus, \(\color{brown}{\log_2(2)=1}\)
SolomonZelman
  • SolomonZelman
So: \(\large\color{black}{ \displaystyle \log_2\left[8\right]=\log_2\left[2^3\right]=3\times \log_2\left[2\right]=3\times 1 =3}\)
SPHott51
  • SPHott51
oh because the log2 2 is same number so it becomes 1..

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