need help with log prob. (1 Viewer)

[PG5]

hey, could anyone please solve this...

2log(base 5)x - 9log(base x)5 = 3

i don't know how to solve this... could ne plz help?

 

[asl2]

1.
log(base5)x(squared) - log(basex)5(squared) = 3

2.
log(base5)x(squared) - log(basex)25 = 3

3.
logx(squared) ............................. log25
___________ ............ (minus) __________ = 3

log5 ............. .. ................ logx


4.
(logx)(logx(squared) - (log5)(log25)
____________________________ = 3

(log5)(logx)



hmmm..... now i've lost it as well

 

[spice girl]

Originally posted by PG5
hey, could anyone please solve this...

2log(base 5)x - 9log(base x)5 = 3

i don't know how to solve this... could ne plz help?

It's a change-of-base problem, based on the rule: log(base y)x = logx/logy

So 2log(base 5)x - 9log(base x)5 = 3

2logx/log5 - 9log5/logx = 3

Use substitution y = logx

2y^2 / log5 - 9log5 = 3y
(2y + 3log5)(y/log5 - 3) = 0

y = 3log5, -(3log5)/2

logx = 3log5, -(3log5)/2
x = 5^3, 5^(-3/2)

 

[PG5]

Re: Re: need help with log prob.

ohh i get it now!

Originally posted by spice girl

2logx/log5 - 9log5/logx = 3

u know how u used the change of base rule on this.... what are the bases for the logs now?
base e? base x?...?

2y^2 / log5 - 9log5 = 3y
(2y + 3log5)(y/log5 - 3) = 0

could u plz provide more calculations b/w these lines? i dun quite get the factorization happening here...

btw, ty for ur help!

 

[wogboy]

u know how u used the change of base rule on this.... what are the bases for the logs now?

Any base is acceptable for the change of base rule (i.e. log(x)/log(n) is always the same value regardless of whatever base the log is taken to), so that would mean for instance:

log(base e)5 / log(base e)3 = log(base 10)5 / log(base 10)3

= log(base x)5 / log(base x) 3, for whatever value x is. Try it with a calculator.

(By convention, ln (which is log to the base of e) is considered the natural logarithm, so that's the log base you will most likely see in textbooks when they talk about the change of base rule).

could u plz provide more calculations b/w these lines? i dun quite get the factorization happening here...

Maybe it would be easier to apply the quadratic formula to the equation to find y, rather than try to figure out what to factorise.

 

[PG5]

Originally posted by wogboy

Maybe it would be easier to apply the quadratic formula to the equation to find y, rather than try to figure out what to factorise.

i tried applying it to the quadratic formula but i got different answers...

 

[spice girl]

Re: Re: need help with log prob.

Originally posted by spice girl


2y^2 / log5 - 9log5 = 3y
(2y + 3log5)(y/log5 - 3) = 0

2y^2 / log5 - 9log5 = 3y
2y^2 -3ylog5 - 9(log5)^2 = 0

Ok using the quadratic formula this time:

y = (1/4)[3log5 +- sqrt(9(log5)^2 + 72(log5)^2)]
= (1/4)[3log5 +- 9log5]
= (1/4)[-6log5], (1/4)[12log5]
= -(3/2)log5, 3log5