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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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I know that log base 25 is actually 1/2 log base 5 y
sure, first you need to know this property of log, \[\log_ba= \dfrac{\log a}{\log b}\]
understood the property

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Other answers:

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

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