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Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$

I have tried to fiddle with it as follows:

$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$ $$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$ Dividing both sides by $6$ gives us $$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$

Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.

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  • $\begingroup$ Please, make titles more informative and less subjective. $\endgroup$ – Pedro Tamaroff May 8 '13 at 1:00
  • $\begingroup$ Ok, I'll make sure I do that in future $\endgroup$ – John Marty May 8 '13 at 1:58
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Let us put $2^x=a,3^x=b$ to remove the indices to improve clarity

So, $8^x=(2^3)^x=(2^x)^3=a^3$ and similarly, $27^x=b^3$

$12^x=(2^2\cdot3)^x=(2^x)^2\cdot3^x=a^2b$ and similarly, $18^x=ab^2$

So, the problem reduces to $$\frac{a^3+b^3}{ab(a+b)}=\frac76$$

$\displaystyle\implies 6(a^2-ab+b^2)=7ab$ as $a+b\ne0$

$\displaystyle\implies 6\left(\frac ab\right)^2-13\cdot\frac ab+6=0$

$\displaystyle\implies \frac ab=\frac32$ or $\dfrac23$

So, $\displaystyle\left(\frac23\right)^x=\frac32$ or $\dfrac23$

If $\displaystyle\left(\frac23\right)^x=\frac32\implies\left(\frac23\right)^x=\left(\frac23\right)^{-1}\iff\left(\frac23\right)^{x+1}=1$

Similarly, if $\displaystyle\left(\frac23\right)^x=\frac23, \left(\frac23\right)^{x-1}=1 $

Now if $\displaystyle u^m=1,$

either $\displaystyle m=0,u\ne0; $

or $\displaystyle u=1$

or $\displaystyle u=-1,m$ is even

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  • $\begingroup$ Thanks so much that's really neat! $\endgroup$ – John Marty May 7 '13 at 4:34
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    $\begingroup$ @JohnMarty, my pleasure. Replacement of numbers with indices by algebraic symbol sometimes helps clarity $\endgroup$ – lab bhattacharjee May 7 '13 at 4:41
  • $\begingroup$ Yeah, when I tried it it looked pretty bewildering, your method simplifies it immeasurably :) $\endgroup$ – John Marty May 7 '13 at 4:42
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    $\begingroup$ @lab-bhattacharjee, I'd say your comment ("Replacement of numbers with indices by algebraic symbol sometimes helps clarity") would go well at the top of your reply. It would help to give context to the solution method, making it less of a 'magical' thing and more of a 'technique' $\endgroup$ – Rolazaro Azeveires May 7 '13 at 8:04
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    $\begingroup$ @RolazaroAzeveires, how about the edited version? $\endgroup$ – lab bhattacharjee May 7 '13 at 8:39
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$$ \begin{array}{rcl} 6(8^x + 27^x) &=& 7(12^x + 18^x) \\ 6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\ \end{array} $$

Substitute $a\!=\!2^x$ and $b\!=\!3^x$ for simplicity:

$$ \begin{array}{rcl} 6(a^3 + b^3) &=& 7(a^2b + ab^2) \\ 6(a+b)(a^2 - ab + b^2) &=& 7ab(a+b) \\ 6(a^2 - ab + b^2) &=& 7ab \\ 6a^2 -13ab + 6b^2 &=& 0 \\ a^2 -\frac{13}{6}ab + b^2 &=& 0 \end{array} $$

The left hand side can be factorized as $\left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)$.

$$ \left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)=0 \\ \begin{array}{rclcrcl} 2a &=& 3b &\text{or}& 3a &=& 2b \\ 2.2^x &=& 3.3^x &\qquad\text{or}\qquad& 3.2^x &=& 2.3^x \\ x &=& -1 &\qquad\text{or}\qquad& x &=& +1 \end{array} $$

Therefore, $x$ can be either $-1$ or $+1$.

$$ \boxed{x = \mp 1} $$

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