Question

Solve algebraically, showing each step of your working, the equation ${{\left( {{8}^{x-1}} \right)}^{2}}-18\left( {{8}^{x-1}} \right)+32=0$.

Answer 1

Hint: o solve the equation we will first substitute ${{8}^{^{x-1}}}=t$ . Now we will get a quadratic equation in t. Now to solve the quadratic equation obtained we will use splitting the middle term method. Hence we will split the middle term such that the product of the terms is the product of the first term and last term. Now we will simplify the equation and hence find the solution of the equation. Now we will substitute the value of t and then solve the equation for x.

Complete step by step solution:
Now let us first consider the given equation ${{\left( {{8}^{x-1}} \right)}^{2}}-18\left( {{8}^{x-1}} \right)+32=0$
Now the given equation looks somewhat like a quadratic equation. So let us convert it into a quadratic equation by substitution. Hence let us substitute $\left( {{8}^{x-1}} \right)=t$ . Now on substituting this we get the equation as, ${{t}^{2}}-18t+32=0$
Now we have a quadratic expression in t. Now we will solve the quadratic equation in t with the help of splitting the middle term method. Hence we will split the middle term such that the product of the terms is same as product of the first term and last term. Hence we can split the middle term as $-18t=-16t-2t$.
Now we get the equation as, ${{t}^{2}}-16t-2t+32=0$
$\begin{align}
  & \Rightarrow t\left( t-16 \right)-2\left( t-16 \right)=0 \\
 & \Rightarrow \left( t-2 \right)\left( t-16 \right)=0 \\
\end{align}$
Now wither $t-2=0$ or $t-16=0$
Hence the value of t is either 2 or 16.
Now let us resubstitute the value of t to find the value of x.
Now let us say $t=2$,
Then we have ${{8}^{x-1}}=2$ .
But this is only possible if power of 8 is $\dfrac{1}{3}$ .
Hence we get, $x-1=\dfrac{1}{3}$
$\begin{align}
  & \Rightarrow x=1+\dfrac{1}{3} \\
 & \Rightarrow x=\dfrac{4}{3} \\
\end{align}$
Now let us say $t=16$
Then we have ${{8}^{x-1}}=16$
Now let us take log on both sides. Hence we get,
$\begin{align}
  & \Rightarrow \log \left( {{8}^{x-1}} \right)=\log 16 \\
 & \Rightarrow x-1\log \left( 8 \right)=\log \left( 16 \right) \\
 & \Rightarrow x-1=\dfrac{\log 16}{\log 8} \\
 & \Rightarrow x-1=\dfrac{\log {{2}^{4}}}{\log {{2}^{3}}} \\
 & \Rightarrow x-1=\dfrac{4\log 2}{3\log 2} \\
 & \Rightarrow x-1=\dfrac{4}{3} \\
\end{align}$
$\begin{align}
  & \Rightarrow x=\dfrac{4}{3}+1 \\
 & \Rightarrow x=\dfrac{4+3}{3} \\
 & \Rightarrow x=\dfrac{7}{3} \\
\end{align}$

Hence the value of x is $\dfrac{7}{3}$ and $\dfrac{4}{3}$.

Note: Now note that whenever we have something in the form of quadratic but not a quadratic equation then we can always use the substitution and solve the equation. Also whenever we are dealing with exponent’s logarithms can be used to simplify the equation.