Question

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.

Answer 1

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

Answer 2

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

Answer 3

understood the property

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

2 log 5 sir

Answer 8

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

Answer 9

i dont know

Answer 10

can you give me the full step solution lol

Answer 11

i am getting confuse here

Answer 12

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

Answer 13

where x=log y

Answer 14

keep it going

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

20 log 5 means 20 log base 5?

Answer 19

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)

Answer 20

i dont get it

Answer 21

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

Answer 22

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

Answer 23

yes

Answer 24

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

Answer 25

okay so there is only 1 value for x?

Answer 26

yes, and so 1 value of y.

Answer 27

so what is it?

Answer 28

i want to see whether i got the right answer.