5logxbase4+48log4basex=x/8
--->
5 log 4 x + 48log x 4 = x/8
One of rules of logs:
log a b = 1/log b a
and furthermore:
log a b = log b/log a
so
5 log 4 x + 48log x 4 = x/8
I don't know of any nice method to solve this, I ran it through Wolfram and got x = 2130.13 , which satisfies the original equation
http://www.wolframalpha.com/input/?i=solve+5(log4)%2Flogx+%2B+48logx%2Flog4+%3D+x%2F8
5logxbase4+48log4basex=x/8
help me plz step
4 answers
My equation near the end should have included:
5 log 4 x + 48log x 4 = x/8
5 log4 x + 48/log4 x = x/2
or
5log 4/log x + 48log x/log 4 = x/2
btw, that is what I used in Wolfram , so my answer stands as is
5 log 4 x + 48log x 4 = x/8
5 log4 x + 48/log4 x = x/2
or
5log 4/log x + 48log x/log 4 = x/2
btw, that is what I used in Wolfram , so my answer stands as is
If we let
u = log_4(x) then x = 4^u and we have
5u + 48/u = 4^u/8
the solution to this is u=4
so, x=4^u = 256
u = log_4(x) then x = 4^u and we have
5u + 48/u = 4^u/8
the solution to this is u=4
so, x=4^u = 256
Steve how did you get u=4?