Question

Determine at which x, the $6$th term in the expansion of the binomial ${(\sqrt{2^{log(10-3^x)}}+\sqrt[5]{52^{(x-2)log3}})}^m$ is equal to $21$, if it is known that the binomial coefficient of the $2$nd, $3$rd and $4$th term in the expansion represent respectively the $1$st, $3$rd and $5$th term of an A.P. (the symbol log stands for logarithm to the base $10$).

Answer 1

$\begin{array}{cc}AIQ,& \\ 2.{}^m{c}_2={}^m{c}_1+{}^m{c}_2& \\ 2.\frac{m\left(m-1\right)}{2}=m+\frac{m\left(m-1\right)\left(m-2\right)}{6}& \\ \left(m-1\right)=1+\frac{\left(m-1\right)\left(m-2\right)}{6}& \\ =6m-12=m^2-3m+2& \\ =m^2-9m+14=0\Rightarrow \left(m-7\right)\left(m-2\right)=0& \\ m=7,m\ne 2& \left[Becauseweseethethirdvalue\right]\\ So,& \\ T_6={}^7{c}_5\left(2^{\log \left(10-3^x\right)}\right)\left(2^{\left(n-2\right)\log 3}\right)=21& \\ T_6={21.2}^{\left[\log \left(10-3^x\right)+\log 3^{\left(x-2\right)}\right]}=21& \left[{}^7{c}_5=21\right]\\ \log \left(10-3^x\right)\left(3^{\left(x-2\right)}\right)=0& \\ \left(10-3^x\right)\left(3^{\left(x-1\right)}\right)=1& \\ {10.3}^{x-2}-3^{2x-2}=1& \\ \frac{{10.3}^x}{3}-\frac{3^{2x}}{3}=1& \\ 3^{2x}-{10.3}^x+9=0\Rightarrow 3^{2x}-{9.3}^x-3^x+9=0& \\ 3^x\left(3^x-9\right)-1\left(3^x-9\right)=0& \\ \left(3^x-1\right)\left(3^x-9\right)=0& \\ 3^x-1\Rightarrow x=0& \\ 3^x-9=0\Rightarrow 3^x=9& \\ x=2& \end{array}$