Question

Determine the value of x for which the 6th term in $(2^{lo{g}_2\left(\sqrt{g^{x-y}+7}\right)}+\frac{1}{2^{\frac{1}{5}lo{g}_2(3^{x-1}+1)}})$ is 84.

Answer 1

Use formula $\Rightarrow a^{Log\begin{array}{c}c\\ a\end{array}}=c^{Log\begin{array}{c}a\\ a\end{array}}$

$\begin{array}{c}\Rightarrow Lo{g}_2\sqrt{9^{x-1}+7}={\left(\sqrt{9^{x-1}+7}\right)}^{Log\begin{array}{c}a\\ a\end{array}}\\ \Rightarrow {\left(\sqrt{9^{x-1}+7}\right)}^1--------\left(i\right)\end{array}$

Use formula $\Rightarrow \log a_a=1$

Now,

$2^{1/5Lo{g}_2\left(3^{x-1}+1\right)}=2^{Lo{g}_2{\left(3^{x-1}+1\right)}^{1/5}}$

Use formula $a^{Log{c}_a=c}$

$\Rightarrow {\left(3^{x-1}+1\right)}^{1/5}--------\left(ii\right)$

$From\left(i\right)\&\left(ii\right)\Rightarrow \left(\sqrt{9^{x-1}+7}+\frac{1}{{\left(3^{x-1}+1\right)}^{1/5}}\right)$

$T_{r+1}{=}^7{c}_r{\left(9^{x-1}+7\right)}^{7-r/2}\times \frac{1}{{\left(3^{x-1}+1\right)}^{r/5}}$

$\begin{array}{c}\because T_r=84\\ \Rightarrow T_{5+1}=84\\ \Rightarrow r=5\\ {\Rightarrow }^7{c}_5{\left(9^{x-1}+7\right)}^1\times \frac{1}{{\left(3^{x-1}+1\right)}^1}=84\end{array}$

Using the formula ${}^n{c}_r=\frac{n!}{n-1!r!}$

$\begin{array}{c}\Rightarrow \left(\frac{7\cdot 6\cdot 5!}{2!5!}{\left(9^{x-1}+7\right)}^1\times \frac{1}{{\left(3^{x-1}+1\right)}^1}\right)=84\\ \Rightarrow 21\left(9^{x-1}+7\right)=84\left(3^{x-1}+1\right)\\ \Rightarrow 9^{x-1}+7=4\left(3^{x-1}+1\right)\end{array}$

$\begin{array}{c}Put{3}^x=t,9^x=t^2\\ \Rightarrow \frac{t^2}{5}+7=\frac{4t+4}{3}\\ \Rightarrow t^2-12t+27=0\\ \Rightarrow \left(t-9\right)\left(t-3\right)=0\\ t=9ort=3\\ \Rightarrow 3^x=3^{2}or{3}^x=3^1\\ x=1or2\end{array}$