log8x - log (1+ x^1/2) = 2

Question

anonymous · Answer

do you know how to reduced the difference between to logs?

kira_yamato · Answer

Try combining the 2 logarithms together. It helps a lot :)

kira_yamato · Answer

But keep in mind that x > 0

unklerhaukus · Answer

is 8 the base of the first log?

unklerhaukus · Answer

or are both logs to base 10?

anonymous · Answer

by division?
would it be 8x/(1+x^1/2)=100. if its so. what next?
What do I do next!
yes

anonymous · Answer

both are base 10

kira_yamato · Answer

Get rid of the denominator.

anonymous · Answer

how?

kira_yamato · Answer

Multiply the equation throughout by the denominator

anonymous · Answer

I did that! seems to make it more complicated.

kira_yamato · Answer

Substitute y = sqrt(x) then ^^

anonymous · Answer

ok will do

unklerhaukus · Answer

another way  from  

here    8x/(1+x^1/2)=100

would be to rationalise the denominator 

( 8x/(1+x^1/2) )×( (1-x^1/2)/(1-x^1/2) )=100

anonymous · Answer

I did rationalize the denominator and that's the point at which i was stuck. the result after rationalizing it was 8x -8x (x^1/2)/ (1-x) = 100

anonymous · Answer

This is the point I got to  8x-8x(x^1/2) = 100 - 100x
108x -8x (x^1/2) = 100 . 
How do i deal with the ever present (x^1/2) ?

kira_yamato · Answer

You could just substitute y as x^1/2 i.e. x = y^2

anonymous · Answer

$$108x -8x(\sqrt{x}) = 100$$

kira_yamato · Answer

$$108y^2 - 8y^3 = 100$$

anonymous · Answer

Currently working using that now.

unklerhaukus · Answer

i dont think there are nice solutions

anonymous · Answer

I don't think so either!

anonymous · Answer

108y^2 -8x^3 =100. This is difficult to solve!!!

anonymous · Answer

mistake 8y^3 not 8x^3

unklerhaukus · Answer

your on the right track,

anonymous · Answer

Thanks to all who tried to help! I will have to get back to it tomorrow!