5(4^(2x-1))=21
4^(2x-1)=21 / 5
=4.2
log[4^(2x-1)]=log4.2
Rule: log (a^x)=x loga
(2x-1) log4 = log4.2
2x-1 = log4.2 / log 4
x= [(log 4.2 / log 4) + 1] / 2
that should be the answer!
I'm having a horrible problem with logarithms...rather natural logarithms. Okay so here's the problem.
5(4^(2x-1))=21
Now I have the soloution, but I need to understand how they got to the solution. There is one other problem, and I'm not sure about it....
But that's a posting for after we work on this one.
PLEASE HELP!!!!
ln21-ln5+ln4
2ln4
here is how to get that
5[4^(2x+1)] = 21
4^(2x+1) = 21/5
ln[4^(2x+1)] = ln (21/5)
Rule:
log(a/b)= loga - logb
(2x+1)ln4 = (ln21 - ln 5)
(2x+1) = (ln21 - ln5) / ln4
2x = (ln21 - ln 5) / ln4 +1= (ln21 - ln5 + ln4) / ln4
x= ln21-ln5+ln4
2ln4
I got confused thinking log was what I needed but if I start with a ln after dividing 21 by 5 then this is the result
ln(4^(2x-1)) = ln(21/5)
pull down (2x-1)
(2x-1)ln4 = ln(21/5)
divide by ln4
2x-1 = ln(21/5)/ln4
separate ln(21/5) into (ln21 - ln5)
2x-1 = (ln21 - ln5)/ln4
add 1
2x = [(ln21-ln5)+ln4]/ln4
divide by 2
x = (ln21-ln5+ln4)/2ln4
f(x) =
inverse function of f(x) is donated by f^-1 (x)
this operation is to isolate f^-1 (x) when x is replaced by f^-1 (x) and f(x) by x in the original function
thus
x = (2 . f^-1 (x) - 3) / ( f^-1 (x) - 5 )
x . f^-1 (x) - 5x = 2 . f^-1 (x) - 3
(x - 2) . f^-1 (x) = -3 + 5x
f^-1 (x) = (-3 + 5x) / (x - 2)
This should be rite!