simple equation

For discussing Olympiad Level Algebra (and Inequality) problems
Posts: 20
Joined: Sat Feb 08, 2014 11:29 pm
Location: Naogaon,Bangladesh

simple equation

Unread post by Sidharth » Sun Feb 09, 2014 11:51 pm

if $(8^x+27^x)/(8^x+12^x)=7/6$ then find the value of x.

Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: simple equation

Unread post by Tahmid » Mon Feb 10, 2014 8:59 pm

i am confused about my solution. plz anyone check it :?:

$\Leftrightarrow 7*(8^{x}+12^{x})=6*(8^{x}+27^{x})$
$\Leftrightarrow (7*8^{x})+(7*12^{x})=(6*8^{x})+(6*27^{x})$
$\Leftrightarrow 7*8^{x}-6*8^{x}=6*27^{x}-7*12^{x}$
$\Leftrightarrow 8^{x}=(6*3^{x}*9^{x})-(7*3^{x}*4^{x})$
$\Leftrightarrow 8^{x}=3^{x}(6*9^{x}-7*4^{x})$
$\Leftrightarrow 3^{x}\mid 8^{x}$

it is possible for x=0.but this solution does not satisfy the property . so there is no positive integer solution for x. :?:

Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka

Re: simple equation

Unread post by photon » Mon Feb 10, 2014 9:17 pm

This problem asks any value of $x$ . No integer solution is possible , but fraction , irrational solution can be possible .
Try not to become a man of success but rather to become a man of value.-Albert Einstein

Nayeemul Islam Swad
Posts: 22
Joined: Sat Dec 14, 2013 3:28 pm

Re: simple equation

Unread post by Nayeemul Islam Swad » Tue Feb 11, 2014 8:25 pm

Letting $a = 2^{x}$ and $b = 3^{x}:$
$\Leftrightarrow \frac{(2^{x})^{3}+(3^{x})^{3}}{4^{x}(2^{x}+3^{x})}=\frac{7}{6}$
$\Leftrightarrow \frac{a^{3}+b^{3}}{a^{2}(a+b)}=\frac{7}{6}$
$\Leftrightarrow \frac{a^{2}-ab+b^{2}}{a^{2}}=\frac{7}{6}$
$\Leftrightarrow \frac{b^{2}-ab}{a^{2}}=\frac{1}{6}$
$\Leftrightarrow (\frac{b}{a})^{2}-(\frac{b}{a})=\frac{1}{6}$
Now, set $c=(\frac{b}{a}):$
$\Leftrightarrow 6c^{2}-6c-1=0$
$\Rightarrow c=\frac{6+\sqrt{60}}{12}[6-\sqrt{60}$ is negative$]$
$\Rightarrow x=\log_{(\frac{3}{2})} (\frac{6+\sqrt{60}}{12})$
Why so SERIOUS?!??!

Post Reply