anonymous
  • anonymous
How can I simplify this: log10(4)log10(2) Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
log(ab) = log a + log b so log 10(4) log 10(2) = (log10 + log4)(log 10+log2) log 10 = 1 (log10 + log4)(log 10+log2)=(1+log4)(1+log2)
anonymous
  • anonymous
I dont get it. But the original expression was: \[Log _{16}(a)+Log _{4}(a)+Log _{2}(a)\]
anonymous
  • anonymous
I was trying to change the log to a unique base. and I got: \[\log(a)=7Log(16)Log(4)Log(2)/(\log4Log2+Log16*Log2+Log16*Log4)\] I got stacked there

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anonymous
  • anonymous
are you asked to find log(a)?
anonymous
  • anonymous
Im asked to find a
anonymous
  • anonymous
i think the expression isn't complete Log16(a)+Log4(a)+Log2(a) it should be an equation or you can't find a
anonymous
  • anonymous
Yeah I was thinking about that too. Ok thanks
anonymous
  • anonymous
Sorry the expresion was: Log16(a)+Log4(a)+Log2(a) =7
anonymous
  • anonymous
do you think the equation can be solved now?
anonymous
  • anonymous
yes
anonymous
  • anonymous
can you explain it to me. I tried really hard and I got no much further.
anonymous
  • anonymous
log16a.4a.2a = 7 log(2^7)(a^3) = 7 7 equals to log 10^7 so log (2^7)(a^3) = log 10^7 (2^7)(a^3) = 10^7 a^3 = 10^7 / 2^7 = (10/2)^7 a^3 = 5^7 a = 5^(7/3)
anonymous
  • anonymous
i got 2^7 from 16x4x2 16 is 2^4 and 4 is 2^2 therefore 16x4x2 = 2^7
anonymous
  • anonymous
what do you mean by: Log16a.4a.2a = 7 \[Log(16a*4a*2a)\] Is 10 the base? How did you end up with that?
anonymous
  • anonymous
yes , if the base isn't written, it means the base is 10. one of the identities of logarithm is loga + logb = log(ab)
anonymous
  • anonymous
Remember the original expression has different bases. \[Log _{16}(a)+Log _{4}(a)+Log _{2}(a)=7\]
anonymous
  • anonymous
oh sorry, i thought the 16 is inside the logarithm. so the base is 16, 4 and 2? wait a moment i'll try solve it again
anonymous
  • anonymous
Ok no problem
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
ok i got it 16 = 2^4 and 4\[a ^{c}\]= 2^2 so \[\log _{2^4}a + \log _{2^2}a + \log _{2}a = 7\] the identity of logarithm: \[\log _{a^b}c = (1/b)logc\] \[(1/4)\log _{2}a + (1/2)\log _{2}a + \log _{2}a = 7\] \[(7/4)\log _{2}a = 7\] \[\log _{2}a = (4/7) 7 = 4\] we know that if \[\log _{a}b = c \] then \[a ^{c}=b\] \[\log _{2}a = (4/7) 7 = 4\] therefore \[2^{4} = a\] a=16
anonymous
  • anonymous
dont mind anything i type above the sentence " the identity of logarithm", it's typo
anonymous
  • anonymous
yeah i was thinking about that. Let me check it out and try to understand it. Thanks dude
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
\[\log_{a ^{b}}x ^{y}=(y/b)\log_{a}x \] put 16 and 4 as a power of 2..and use the above log property. hope this would help
anonymous
  • anonymous
yeah thanks. It is clear now

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