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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
    May 8, 2013 at 1:00
  • $\begingroup$ Ok, I'll make sure I do that in future $\endgroup$
    – John Marty
    May 8, 2013 at 1:58

3 Answers 3

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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, 2013 at 4:34
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    $\begingroup$ @JohnMarty, my pleasure. Replacement of numbers with indices by algebraic symbol sometimes helps clarity $\endgroup$ May 7, 2013 at 4:41
  • $\begingroup$ Yeah, when I tried it it looked pretty bewildering, your method simplifies it immeasurably :) $\endgroup$
    – John Marty
    May 7, 2013 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' $\endgroup$ May 7, 2013 at 8:04
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    $\begingroup$ @RolazaroAzeveires, how about the edited version? $\endgroup$ May 7, 2013 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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$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$

Divide by$12^x$

$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$

$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$

We put :$t=(\frac{3}{2}) ^x$

$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$

Multiply by $t$

$\Rightarrow $ $ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$

We can see $-1$ is a solution of the equation

So:after division by $t+1$ we see

$ t^2 - \frac{13}{6}x+1=0$

$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $

So :

$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $

$\Rightarrow $ $x_1=1$

And

$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $

$\Rightarrow $ $x_2=-1$

Finally

$x=-1$ or $ x= 1$

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