anonymous
  • anonymous
3 log base 5 y - 1og base 25 y = 10 This is a difficult question for me to solve and i hope you would help me thanks.
Mathematics
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anonymous
  • anonymous
3 log base 5 y - 1og base 25 y = 10 This is a difficult question for me to solve and i hope you would help me thanks.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I know that log base 25 is actually 1/2 log base 5 y
hartnn
  • hartnn
sure, first you need to know this property of log, \[\log_ba= \dfrac{\log a}{\log b}\]
anonymous
  • anonymous
understood the property

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hartnn
  • hartnn
so, \(\log_5 y= ...?? \\ \log_{25}y=... ?\)
anonymous
  • anonymous
log base 5 y = log y / log 5 and log base 25 y = log y/ log 25
hartnn
  • hartnn
good :) now \(\log x^n=n \log x \\ so, \log 25=.... ?\)
anonymous
  • anonymous
2 log 5 sir
hartnn
  • hartnn
so, now you can put log y =x and the equation will become ...?
anonymous
  • anonymous
i dont know
anonymous
  • anonymous
can you give me the full step solution lol
anonymous
  • anonymous
i am getting confuse here
hartnn
  • hartnn
\(3x/log 5 -x/(2log 5) = 10 \\ \) got this step first ?
hartnn
  • hartnn
where x=log y
anonymous
  • anonymous
keep it going
hartnn
  • hartnn
multiply both sides by 2log 5 6x - x = 20 log 5 5x = 20 log 5 got this much ?
anonymous
  • anonymous
The first step that you posted is that the first step of showing the workings?
hartnn
  • hartnn
if you understand how to solve, you will be able to show the workings on your own....
anonymous
  • anonymous
20 log 5 means 20 log base 5?
hartnn
  • hartnn
i have made the base unspecified (when its unspecified, by default its 10), so 20 log 5 means 20 times log of 5 (with the base 10)
anonymous
  • anonymous
i dont get it
anonymous
  • anonymous
i think that is where the problem lies in. if not, i could easily understand your workings.
hartnn
  • hartnn
\(\log_ba= \dfrac{\log_{10} a}{\log_{10} b} \\\log _{25}y =\dfrac{\log_{10} y}{\log_{10} 25} \) now ?
anonymous
  • anonymous
yes
hartnn
  • hartnn
\(\log_ba= \dfrac{\log_{10} a}{\log_{10} b} \\\log _{25}y =\dfrac{\log_{10} y}{\log_{10} 25}=\dfrac{\log_{10} y}{2\log_{10} 5}\\\log_{5}y=\dfrac{\log_{10} y}{\log_{10} 5} \\ let \log y=x \\ 3x/\log 5 -x/(2\log 5)=10 \implies 6x-x=20\log_{10}5\\x=4\log_{10}5 \\ \log_{10} y = \log _{10}5^4 \implies y= 5^4\)
anonymous
  • anonymous
okay so there is only 1 value for x?
hartnn
  • hartnn
yes, and so 1 value of y.
anonymous
  • anonymous
so what is it?
anonymous
  • anonymous
i want to see whether i got the right answer.

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