anonymous
  • anonymous
log 7 √ 49
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
first off, \(\sqrt{49}=7\) right?
anonymous
  • anonymous
\[\sqrt{7} 49\]
anonymous
  • anonymous
so you are being asked for \[\log_7(7)\] which must be 1, because you are being asked "raise 7 to what power to get 7?" and the answer is 1

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anonymous
  • anonymous
now i am lost
anonymous
  • anonymous
what is the base of your log? what is the input?
amorfide
  • amorfide
if you have this as base 10 then you are multiplying 7 by 7=49 log49 but if it is base 7, follow what @satellite73 said
anonymous
  • anonymous
log sqrt 7 bqase 49
anonymous
  • anonymous
Base**
amorfide
  • amorfide
lol what?
anonymous
  • anonymous
\[\log_{49}(\sqrt{7})\]?
anonymous
  • anonymous
base is 49, input is \(\sqrt{7}\) right?
anonymous
  • anonymous
trick is to try to write with exponents, because \[\log_b(x)=y\iff b^y=x\] so you are trying to solve \[49^y=\sqrt{7}\]
anonymous
  • anonymous
we see that \(49=7^2\) and \(\sqrt{7}=7^{\frac{1}{2}}\) so what you want is \[7^{2y}=7^{\frac{1}{2}}\] or \[2y=\frac{1}{2}\] or \[y=\frac{1}{4}\]
anonymous
  • anonymous
meaning is simpler english that the number \(\sqrt{7}\) is the fourth root of 49

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