Algebra Problem of the Day-2

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

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

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 \(\displaystyle x=-1\)
Click to expand...
Well done! ???

As for Subhotosh Khan, you'll get a participant medal for your paltry attempt. Maybe next time. ?
 
Harry_the_cat said:
Letting \(\displaystyle a=2^x\) and \(\displaystyle b=3^x\) for ease of typing!!
Click to expand...
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!
 
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
 
Steven G said:
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
Click to expand...
Yeah sure, but does that tell you that they are the ONLY solutions?
 
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