Question

Number of values of x for which $\frac{8^x+{27}^x}{{12}^x+{18}^x}=\frac{7}{6}$

• A. $2$
• B. $3$
• C. $1$
• D. No value of x

Answer 1

The correct option is A $2$

Given: $\frac{8^x+{27}^x}{{12}^x+{18}^x}=\frac{7}{6}$

To find number of values of x

Sol: Let $2^x=a,3^x=b$

So $8^x=(2^3{)}^x=(2^x{)}^3=a^3$ similarly ${27}^x=b^3$

and ${12}^x=(2^2.3{)}^x=(2^x{)}^2{.3}^x=a^{2}b$ simialrly ${18}^x=({2.3}^2{)}^x=2^x.(3^x{)}^2=a{b}^2$

So the given problem becomes,

$\frac{a^3+b^3}{a^{2}b+a{b}^2}=\frac{7}{6}⟹\frac{(a+b)(a^2–ab+b^2)}{ab(a+b)}=\frac{7}{6}⟹\frac{a^2-ab+b^2}{ab}=\frac{7}{6}⟹6a^2-6ab+6b^2=7ab⟹6a^2-13ab+6b^2=0$

As $a+b\ne 0$. Divide throughout by $b^2$

$⟹6{\left(\frac{a}{b}\right)}^2-13\left(\frac{a}{b}\right)+6=0$

Let $\frac{a}{b}=z$

$6z^2-13z+6=0⟹(2z-3)(3z-2)=0⟹z=\frac{2}{3},z=\frac{3}{2}$

or $\frac{a}{b}=\frac{2}{3}$ or $\frac{3}{2}$

or ${\left(\frac{2}{3}\right)}^x=\frac{2}{3}$ or $\frac{3}{2}$

If ${\left(\frac{2}{3}\right)}^x=\frac{2}{3}⟹{\left(\frac{2}{3}\right)}^x={\left(\frac{2}{3}\right)}^1⟹{\left(\frac{2}{3}\right)}^{x-1}=1$

and if ${\left(\frac{2}{3}\right)}^x=\frac{3}{2}⟹{\left(\frac{2}{3}\right)}^x={\left(\frac{2}{3}\right)}^{-1}⟹{\left(\frac{2}{3}\right)}^{x+1}=1$

That means $x$ can either be $-1$ or $+1$