Sgstudent
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.



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Sgstudent
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I know that log base 25 is actually 1/2 log base 5 y

hartnn
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sure, first you need to know this property of log,
\[\log_ba= \dfrac{\log a}{\log b}\]

Sgstudent
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understood the property

hartnn
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so, \(\log_5 y= ...?? \\ \log_{25}y=... ?\)

Sgstudent
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log base 5 y = log y / log 5 and log base 25 y = log y/ log 25

hartnn
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good :)
now \(\log x^n=n \log x \\ so, \log 25=.... ?\)

Sgstudent
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2 log 5 sir

hartnn
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so, now you can put log y =x
and the equation will become ...?

Sgstudent
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i dont know

Sgstudent
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can you give me the full step solution lol

Sgstudent
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i am getting confuse here

hartnn
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\(3x/log 5 x/(2log 5) = 10 \\ \)
got this step first ?

hartnn
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where x=log y

Sgstudent
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keep it going

hartnn
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multiply both sides by 2log 5
6x  x = 20 log 5
5x = 20 log 5
got this much ?

Sgstudent
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The first step that you posted is that the first step of showing the workings?

hartnn
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if you understand how to solve, you will be able to show the workings on your own....

Sgstudent
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20 log 5 means 20 log base 5?

hartnn
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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)

Sgstudent
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i dont get it

Sgstudent
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i think that is where the problem lies in. if not, i could easily understand your workings.

hartnn
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\(\log_ba= \dfrac{\log_{10} a}{\log_{10} b} \\\log _{25}y =\dfrac{\log_{10} y}{\log_{10} 25} \)
now ?

Sgstudent
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yes

hartnn
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\(\log_ba= \dfrac{\log_{10} a}{\log_{10} b} \\\log _{25}y =\dfrac{\log_{10} y}{\log_{10} 25}=\dfrac{\log_{10} y}{2\log_{10} 5}\\\log_{5}y=\dfrac{\log_{10} y}{\log_{10} 5} \\ let \log y=x \\ 3x/\log 5 x/(2\log 5)=10 \implies 6xx=20\log_{10}5\\x=4\log_{10}5 \\ \log_{10} y = \log _{10}5^4 \implies y= 5^4\)

Sgstudent
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okay so there is only 1 value for x?

hartnn
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yes, and so 1 value of y.

Sgstudent
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so what is it?

Sgstudent
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i want to see whether i got the right answer.