Algebra Problem of the Day-2

[BigBeachBanana]

$$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$Find $x$.

 

[Deleted member 4993]

$$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$Find $x$.

............↑........right there

 

[Harry_the_cat]

$\displaystyle \frac{(2^x)^3+(3^x)^3}{2^x3^x(2^x+3^x)} = \frac{7}{6}$

Letting $\displaystyle a=2^x$ and $\displaystyle b=3^x$ for ease of typing!!

$\displaystyle \frac{(a^3+b^3)}{ab(a+b)} =\frac{7}{6}$

$\displaystyle \frac{(a+b)(a^2-ab+b^2)}{ab(a+b)} = \frac{7}{6}$

$\displaystyle \frac{a^2-ab+b^2}{ab} = \frac{7}{6}$ since $\displaystyle a+b\neq0$

$\displaystyle 6a^2 -6ab +6b^2 =7ab$

$\displaystyle 6a^2 - 13ab+6b^2=0$

$\displaystyle (3a-2b)(2a-3b)=0$

$\displaystyle 3a-2b=0$ OR $\displaystyle 2a-3b=0$ OR $\displaystyle a=b=0$ but $\displaystyle a\neq0$ and $\displaystyle b\neq0$

$\displaystyle \frac{a}{b}=\frac{2}{3}$ OR $\displaystyle \frac{a}{b}=\frac{3}{2}$

$\displaystyle \frac{2^x}{3^x} =\frac{2}{3}$ OR $\displaystyle \frac{2^x}{3^x}=\frac{3}{2}$

$\displaystyle (\frac{2}{3})^x =\frac{2}{3}$ OR $\displaystyle (\frac{2}{3})^x=\frac{3}{2}$

$\displaystyle x=1$ OR $\displaystyle x=-1$

 

[Harry_the_cat]

............↑........right there

Nice!

 

[BigBeachBanana]

Spoiler

$\displaystyle \frac{(2^x)^3+(3^x)^3}{2^x3^x(2^x+3^x)} = \frac{7}{6}$

Letting $\displaystyle a=2^x$ and $\displaystyle b=3^x$ for ease of typing!!

$\displaystyle \frac{(a^3+b^3)}{ab(a+b)} =\frac{7}{6}$

$\displaystyle \frac{(a+b)(a^2-ab+b^2)}{ab(a+b)} = \frac{7}{6}$

$\displaystyle \frac{a^2-ab+b^2}{ab} = \frac{7}{6}$ since $\displaystyle a+b\neq0$

$\displaystyle 6a^2 -6ab +6b^2 =7ab$

$\displaystyle 6a^2 - 13ab+6b^2=0$

$\displaystyle (3a-2b)(2a-3b)=0$

$\displaystyle 3a-2b=0$ OR $\displaystyle 2a-3b=0$ OR $\displaystyle a=b=0$ but $\displaystyle a\neq0$ and $\displaystyle b\neq0$

$\displaystyle \frac{a}{b}=\frac{2}{3}$ OR $\displaystyle \frac{a}{b}=\frac{3}{2}$

$\displaystyle \frac{2^x}{3^x} =\frac{2}{3}$ OR $\displaystyle \frac{2^x}{3^x}=\frac{3}{2}$

$\displaystyle (\frac{2}{3})^x =\frac{2}{3}$ OR $\displaystyle (\frac{2}{3})^x=\frac{3}{2}$

$\displaystyle x=1$ OR $\displaystyle x=-1$

Well done! ???

---

As for Subhotosh Khan, you'll get a participant medal for your paltry attempt. Maybe next time. ?

 

[Cubist]

Letting $\displaystyle a=2^x$ and $\displaystyle b=3^x$ for ease of typing!!

I was about to post my solution. However, I won't bother after seeing yours since it's pretty identical to mine. And you beat me by 45 mins - also I didn't make the "ease of typing" substitution - brilliant!

 

[Steven G]

It appears that making the substitution actually made this problem harder (in this case)

$\displaystyle \dfrac{8^x +27^x}{12^x + 18^x}=\dfrac{(2x)^3 + (3x)^3}{6^x(2^x + 3^x)} = \dfrac{(2^x)^2 -(2^x)(3^x) +(3^x)^2}{2^x3^x} =(2/3)^x - 1 +(3/2)^x = \dfrac{7}{6} or (2/3)^x +(3/2)^x = \dfrac{13}{6}$

It is easily seen (luckily!) that since 2/3 + 3/2 = 13/6 the answers are x=1 or -1

 

[Harry_the_cat]

It appears that making the substitution actually made this problem harder (in this case)

$\displaystyle \dfrac{8^x +27^x}{12^x + 18^x}=\dfrac{(2x)^3 + (3x)^3}{6^x(2^x + 3^x)} = \dfrac{(2^x)^2 -(2^x)(3^x) +(3^x)^2}{2^x3^x} =(2/3)^x - 1 +(3/2)^x = \dfrac{7}{6} or (2/3)^x +(3/2)^x = \dfrac{13}{6}$

It is easily seen (luckily!) that since 2/3 + 3/2 = 13/6 the answers are x=1 or -1

Yeah sure, but does that tell you that they are the ONLY solutions?

 

[Steven G]

Yeah sure, but does that tell you that they are the ONLY solutions?

I was waiting for that comment. I guess that I should have mentioned that I saw the quadratic coming up and hence did not expect more than two answers.