# Thread: need help with some tricky log equation

1. ## need help with some tricky log equation

Hi all! I am a bit new here.
I am more a programmer and not really strong at mathematics. One day I noticed some advertisement campaign which asks to solve some equation to become a participant of a lottery. The organizer is pretty serious about that and is well known solid company in my country. So I decided to give it a try.
But after trying to solve the eq I got stuck at some point.
I have attached the png image from the advertisement leaflet with this equation.
Now I'll tell you how far did I get.
BTW, I am using Maple to help me with getting everything done nicely But Maple is not able to do everything for me.

So, here it goes, the original eq in Maple:
log[2](5*sqrt((3^(-4)*x+10)*1/24-30)-sqrt((x+36)*1/78-191)) = (1+tg(x)^2)/sin(x)*(1-cos(2*x)^2)/sin(x)-(2+cos(x)^2)/log[2](10+x^2)

(in Maple notation log[2] means log with base 2)

At first I tried to simplify everything.
This one
(1+tg(x)^2)/sin(x)*(1-cos(2*x)^2)/sin(x)
using the law 1+tg(x)^2 = 1/cos(x)^2
and
1 = cos(2*x)^2+sin(2*x)^2
and
sin(2*x) = 2sin(x)*cos(x) => sin(2*x)^2 = 4*sin(x)^2*cos(x)^2

easily transformed into simply number 4

5*sqrt((3^(-4)*x+10)*1/24-30)-sqrt((x+36)*1/78-191)
transformed into
(5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236)
(nothing useful at this moment )

So the right side is
4-(2+cos(x)^2)/log[2](10+x^2)

somehow Maple decided to change the base of log[2] and transformed it into ln. Now it looked like this:
4-(2+cos(x)^2)*ln(2)/ln(10+x^2)

I decided to give it a try and transform everything so I have lns on both sides. So after a while I got the right side:
(4*ln(10+x^2)-(2+cos(x)^2)*ln(2))/ln(10+x^2) => (using log coefficient property) => (ln(10+x^2)^4-ln(2^(2+cos(x)^2)))/ln(10+x^2) => (using log subtraction porperty) => [IMG]file:///C:/DOCUME%7E1/progmars/LOCALS%7E1/Temp/moz-screenshot-13.jpg[/IMG]ln((10+x^2)^4/2^(2+cos(x)^2))/ln(10+x^2)

And transforming the left side to ln and putting the right side to left:
ln((5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236))/ln(2)-ln((10+x^2)^4/2^(2+cos(x)^2))/ln(10+x^2) = 0

Trying to get rid of fractions got this:
ln(10+x^2)*ln((5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236))-ln((10+x^2)^4/2^(2+cos(x)^2))*ln(2) = 0

and using log coefficient property again:
ln(((5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236))^ln(10+x^2)-ln(((10+x^2)^4/2^(2+cos(x)^2))^ln(2))) = 0

using log subtraction porperty again:
ln(((5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236))^ln(10+x^2)/((10+x^2)^4/2^(2+cos(x)^2))^ln(2)) = 0

Now it is obvius, that if I have ln(x)=0; then x=1:
((5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236))^ln(10+x^2)/((10+x^2)^4/2^(2+cos(x)^2))^ln(2) = 1
or another way:
((5/108)*sqrt(6*x-345060)-(1/78)*sqrt(78*x-1159236))^ln(10+x^2) = ((10+x^2)^4/2^(2+cos(x)^2))^ln(2)

So I got an eq in a form f(x)^g(x) = h(x)^ln(2)
The question is - what to do next? Will it get simpler or maybe I did it the wrong way from the very beginning?

I did try to give Maple a plot command and it was obvious that there ARE multiple solutions in the region 7849925599 .. 8009925599. I am pretty sure that it is the right way because there was a hint given in the campaign that I should get a 10 digit number at the end.

Thanks for any ideas!