BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,185
[math]\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}[/math]Find [imath]x[/imath].
............↑........right there[math]\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}[/math]Find [imath]x[/imath].
Nice!............↑........right there
Well done! ???\(\displaystyle \frac{(2^x)^3+(3^x)^3}{2^x3^x(2^x+3^x)} = \frac{7}{6}\)
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 \(\displaystyle x=-1\)
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!Letting \(\displaystyle a=2^x\) and \(\displaystyle b=3^x\) for ease of typing!!
Yeah sure, but does that tell you that they are the ONLY solutions?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
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.Yeah sure, but does that tell you that they are the ONLY solutions?