Question

log base sqrt2 Y = 3 + log base 2(Y+6)...Find Y

Answer 1

3=log base 2 2^3= log base 2 8

Answer 2

=log base 2 8 + log base 2 (Y+6)

=log base 2 (8)(Y+6)

so, Y=8Y+48
and Y= -48/7

Answer 3

\(\huge \log_{\sqrt{2}} \ Y = 3 + \log_2 (Y+6)\)

Answer 4

its like this ?

Answer 5

ganeshie8 yes

Answer 6

ok first we need to bring logs to same base

Answer 7

\(\huge \log_{2^{1/2}} Y = 3 + \log_2 (Y+6)\)

\(\huge \frac{\log_{2} Y}{1/2} \ = 3 + \log_2 (Y+6) \)

\(\huge 2\log_2 Y - \log_2 (Y+6) = 3\)

\(\huge \log_2 Y^2 - \log_2 (Y+6) = 3\)

Answer 8

so far makes sense ?

we have used the log property :

\(\log_{a^m} b = \frac{\log_a b}{m}\)

Answer 9

i wasn't taught this log property. Mine working was \[\log_{\sqrt{2}}Y =\log_{2}Y \div \log_{2}\sqrt{2} \]

Answer 10

hmm we can use this also. but this makes the solution bit messy
its good you learn that property

Answer 11

can u work out the answer for me? I want to tally with mine. thanks

Answer 12

rest of the steps are straight forward.. il put them wait

Answer 13

\(\huge \frac{Y^2}{(Y+6)} = 2^3\)

\(\huge Y^2 = 8Y + 48\)

\(\huge Y^2 - 8Y - 48 = 0\)

=> Y = -4, 12

since Y cannot be negative.

Y = 12

Answer 14

thank you...using my method i get y = -7.06 which is wrong but I don't know the right ans so shall use yours as my correction.

Answer 15

you can plugin 12 and check. it satisfies...

but your method also should give 12

Answer 16

your method is more basic... it should work as well. right now im running... maybe you msg me ur work if psble... il have a look ok :D

Answer 17

I will post the pic later and u check for me when free..thank you very much,

Answer 18

sure np :)

wolf says Y=12 is correct

http://www.wolframalpha.com/input/?i=3+%2B+%28log+18%29%2Flog+2+-+%28log+12%29%2F+log+%28sqrt+2%29