simple equation
if $(8^x+27^x)/(8^x+12^x)=7/6$ then find the value of x.
Re: simple equation
i am confused about my solution. plz anyone check it
$\frac{8^{x}+27^{x}}{8^{x}+12^{x}}=\frac{7}{6}$
$\Leftrightarrow 7*(8^{x}+12^{x})=6*(8^{x}+27^{x})$
$\Leftrightarrow (7*8^{x})+(7*12^{x})=(6*8^{x})+(6*27^{x})$
$\Leftrightarrow 7*8^{x}-6*8^{x}=6*27^{x}-7*12^{x}$
$\Leftrightarrow 8^{x}=(6*3^{x}*9^{x})-(7*3^{x}*4^{x})$
$\Leftrightarrow 8^{x}=3^{x}(6*9^{x}-7*4^{x})$
$\Leftrightarrow 3^{x}\mid 8^{x}$
it is possible for x=0.but this solution does not satisfy the property . so there is no positive integer solution for x.
$\frac{8^{x}+27^{x}}{8^{x}+12^{x}}=\frac{7}{6}$
$\Leftrightarrow 7*(8^{x}+12^{x})=6*(8^{x}+27^{x})$
$\Leftrightarrow (7*8^{x})+(7*12^{x})=(6*8^{x})+(6*27^{x})$
$\Leftrightarrow 7*8^{x}-6*8^{x}=6*27^{x}-7*12^{x}$
$\Leftrightarrow 8^{x}=(6*3^{x}*9^{x})-(7*3^{x}*4^{x})$
$\Leftrightarrow 8^{x}=3^{x}(6*9^{x}-7*4^{x})$
$\Leftrightarrow 3^{x}\mid 8^{x}$
it is possible for x=0.but this solution does not satisfy the property . so there is no positive integer solution for x.
Re: simple equation
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
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Re: simple equation
Letting $a = 2^{x}$ and $b = 3^{x}:$
$\frac{8^{x}+27^{x}}{8^{x}+12^{x}}=\frac{7}{6}$
$\Leftrightarrow \frac{(2^{x})^{3}+(3^{x})^{3}}{4^{x}(2^{x}+3^{x})}=\frac{7}{6}$
$\Leftrightarrow \frac{a^{3}+b^{3}}{a^{2}(a+b)}=\frac{7}{6}$
$\Leftrightarrow \frac{a^{2}-ab+b^{2}}{a^{2}}=\frac{7}{6}$
$\Leftrightarrow \frac{b^{2}-ab}{a^{2}}=\frac{1}{6}$
$\Leftrightarrow (\frac{b}{a})^{2}-(\frac{b}{a})=\frac{1}{6}$
Now, set $c=(\frac{b}{a}):$
$c^{2}-c=\frac{1}{6}$
$\Leftrightarrow 6c^{2}-6c-1=0$
$\Rightarrow c=\frac{6+\sqrt{60}}{12}[6-\sqrt{60}$ is negative$]$
$\Rightarrow(\frac{3}{2})^{x}=\frac{6+\sqrt{60}}{12}$
$\Rightarrow x=\log_{(\frac{3}{2})} (\frac{6+\sqrt{60}}{12})$
$\frac{8^{x}+27^{x}}{8^{x}+12^{x}}=\frac{7}{6}$
$\Leftrightarrow \frac{(2^{x})^{3}+(3^{x})^{3}}{4^{x}(2^{x}+3^{x})}=\frac{7}{6}$
$\Leftrightarrow \frac{a^{3}+b^{3}}{a^{2}(a+b)}=\frac{7}{6}$
$\Leftrightarrow \frac{a^{2}-ab+b^{2}}{a^{2}}=\frac{7}{6}$
$\Leftrightarrow \frac{b^{2}-ab}{a^{2}}=\frac{1}{6}$
$\Leftrightarrow (\frac{b}{a})^{2}-(\frac{b}{a})=\frac{1}{6}$
Now, set $c=(\frac{b}{a}):$
$c^{2}-c=\frac{1}{6}$
$\Leftrightarrow 6c^{2}-6c-1=0$
$\Rightarrow c=\frac{6+\sqrt{60}}{12}[6-\sqrt{60}$ is negative$]$
$\Rightarrow(\frac{3}{2})^{x}=\frac{6+\sqrt{60}}{12}$
$\Rightarrow x=\log_{(\frac{3}{2})} (\frac{6+\sqrt{60}}{12})$
Why so SERIOUS?!??!