Question

If ${\left(\frac{9}{7}\right)}^3\times {\left(\frac{49}{81}\right)}^{(2x-6)}={\left(\frac{7}{9}\right)}^9$ the value of $x$is

Answer 1

Solve for the value of the unknown variable using the exponent's rule.

${\left(\frac{9}{7}\right)}^3\times {\left(\frac{49}{81}\right)}^{(2x-6)}={\left(\frac{7}{9}\right)}^9$

We know that the reciprocal of ${\left(\frac{9}{7}\right)}^3$ is ${\left(\frac{7}{9}\right)}^{-3}$

$$\begin{gathered} \Rightarrow {\left(\frac{49}{81}\right)}^{2x-6}={\left({\left(\frac{7}{9}\right)}^2\right)}^{2x-6} \\ \Rightarrow {\left(\frac{7}{9}\right)}^{-3}\times {\left({\left(\frac{7}{9}\right)}^2\right)}^{(2x-6)}={\left(\frac{7}{9}\right)}^9\left[{\left(a^n\right)}^m=a^{mn}\right] \end{gathered}$$

All the bases are the same, hence equating the powers we get,

$$\begin{gathered} \Rightarrow -3+2(2x–6)=9 \\ \Rightarrow -3+4x–12=9 \\ \Rightarrow 4x=9+15 \\ \Rightarrow 4x=24 \\ \Rightarrow x=6 \end{gathered}$$

Hence, the required answer is $x=6$