Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
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.
$\endgroup$ 23 Answers
$\begingroup$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
$\endgroup$ 6 $\begingroup$$$ \begin{array}{rcl} 6(8^x + 27^x) &=& 7(12^x + 18^x) \\ 6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\ \end{array} $$
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 $+1$.
$$ \boxed{x = \mp 1} $$
$\endgroup$ $\begingroup$$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$
Divide by$12^x$
$\Rightarrow $$ 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$
$\endgroup$
