Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$

Question

Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$

Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$

from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$

since $7$ is a prime we can say that $$a+b=1, a^2-ab+b^2=7.$$

It follows that $$a=1-b$$

from where $$(1-b)^2-(1-b)b+b²=7$$

this quadratic has solutions $b=1, b=0.$

what I now did was consider cases. Firstly $a=b=0$ which has no solutions for $x$.

case $a=b=1$ has the solution $x=0$.

However the actual solutions for this were $x=1, x=-1$ which I don't see how they came up with. What is wrong with my approach?

Answer 1

The problem here is that $a+b$ and $a^2 - ab + b^2$ are not necessarily integers.

@lab bhattacharjee gives a complete answer for the same problem. In short, observe that both $a^3+b^3$ and $a^2b + ab^2$ are homogeneous of degree $3$. Dividing by $a^3$ yields $$\frac{7}{6}=\frac{a^3+b^3}{a^2b+ab^2}=\frac{1+t^3}{t+t^2}=\frac{1-t+t^2}{t}$$ where $t=\frac{b}{a}=\left( \frac 3 2 \right)^x$. You can proceed from here.

Answer 2

$$\frac{a^3 + b^3}{a^2b + ab^2} = \frac{7}{6} \implies 7 a^2b + 7ab^2 = 6 a^3 + 6b^3 \implies 7ab(a + b) = 6(a + b)(a^2 -a b + b^2)$$ $$ \implies (a + b)(6a^2 -13ab + 6b^2) =0 $$ Then either $(a + b) = 0$ or $(6a^2 -13ab + 6b^2) = (2a -3b)(3a - 2b) = 0 $

Hence, we have that $$ a = - b \implies 2^x = - 3^x \implies \left(\frac{2}{3}\right)^x = -1 $$ which is not possible. Hence, $$ 2a = 3b \implies 2^{x+ 1} = 3^{x +1 } \implies \left(\frac{2}{3}\right)^{x + 1} = \left(\frac{2}{3} \right)^0 \implies x = -1$$ Or, $$ 3a = 2b \implies 2^{x- 1} = 3^{x -1 } \implies \left(\frac{2}{3}\right)^{x + 1} = \left(\frac{2}{3} \right)^0 \implies x = 1 $$ Thus, $x = 1, -1$ are the only soluions.

Answer 3

We have that

$$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x+3^x)(2^{2x}-6^x+3^{2x})}{(6^x)(2^x+3^x)}=\frac{2^{2x}-6^x+3^{2x}}{6^x}=\frac76$$

$$\iff\left(\frac23\right)^x+\left(\frac32\right)^x=\frac{13}6$$

$$\iff y+\frac1y=\frac{13}6 \iff y^2-\frac{13}6y+1=0$$

which leads to $y=\frac23,\frac32$ and $x=\pm1$.

Answer 4

You properly wrote $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$ Tak into account the homegeneity and let $b=k a$ to get $$k+\frac 1k-1=\frac{7}{6}\implies k=\frac 23 \qquad \text{and} \qquad k=\frac 32$$ and then the slutions.