47

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.

Cookie
  • 13,532
John Marty
  • 3,650

3 Answers3

68

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

lab bhattacharjee
  • 274,582
  • Thanks so much that's really neat! – John Marty May 07 '13 at 04:34
  • 4
    @JohnMarty, my pleasure. Replacement of numbers with indices by algebraic symbol sometimes helps clarity – lab bhattacharjee May 07 '13 at 04:41
  • Yeah, when I tried it it looked pretty bewildering, your method simplifies it immeasurably :) – John Marty May 07 '13 at 04:42
  • 5
    @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' – Rolazaro Azeveires May 07 '13 at 08:04
  • 1
    @RolazaroAzeveires, how about the edited version? – lab bhattacharjee May 07 '13 at 08:39
  • @lab-bhattacharjee, better I think (2 weeks latter I alredy do not recall the initial version..) I probably would go as far as clarifying upfront that the choice of bases 2 and 3 are because they are the prime factors in 8, 12, and so on; but I guess that is clear enough from the immediate development (and thanks) – Rolazaro Azeveires May 20 '13 at 19:46
6

$$ \begin{array}{rcl} 6(8^x + 27^x) &=& 7(12^x + 18^x) \\ 6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\ \end{array} $$

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 $+1$.

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

hkBattousai
  • 4,543
0

$\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$