Question

How to solve for x? \[8\times {{2}^{2x}}+4\times {{2}^{x+1}}=1+{{2}^{x}}\]

Answer 1

Hint: In the above question we have to follow some exponential rules in order to solve for x. Here we will first simplify the exponents using the exponential rules. Then we have to make some assumptions in order to get a basic quadratic equation. After that, we will solve the quadratic equation using either factorization or by completing the square method. After simplification, we will get the value of.

Complete step by step answer:
This is an exponential equation and to solve it we have two methods. If the equation is already in simplified form, we can directly solve it by applying general exponential rules or logarithmic rules.
In this case first we will first simplify to get a simpler form of equation. Then we will isolate the exponential term to one side
\[\begin{align}
  & 8\times {{2}^{2x}}+4\times {{2}^{x+1}}=1+{{2}^{x}} \\
 & \Rightarrow 8\times {{2}^{2x}}+4\times {{2}^{x+1}}-1-{{2}^{x}}=0 \\
 & \Rightarrow 8\times {{\left( {{2}^{x}} \right)}^{2}}+4\times {{2}^{x}}\times {{2}^{1}}-1-{{2}^{x}}=0 \\
\end{align}\]
Here we can see that the equation is now simplified. Now, we will convert it into quadratic equation which is \[a{{x}^{2}}+bx+c=0\]
Let \[{{2}^{x}}=y\]
Therefore, the equation becomes
\[\begin{align}
  & 8\times {{\left( y \right)}^{2}}+8y-1-y=0 \\
 & \Rightarrow 8{{y}^{2}}+7y-1=0 \\
\end{align}\]
Now simplify the quadratic equation using factorization. First, we have to split the middle term \[bx\] in two terms, such that the product of the two terms is equal to the constant \[c\] and the sum of the two terms is equal to the middle term \[bx\]. Then factor the first two terms and the last two terms.
\[8{{y}^{2}}+7y-1=0\]
\[\begin{align}
  & \Rightarrow 8{{y}^{2}}+8y-y-1=0 \\
 & \Rightarrow 8y(y+1)-1(y+1)=0 \\
 & \Rightarrow (8y-1)(y+1)=0 \\
 & \Rightarrow y=\dfrac{1}{8},y=-1 \\
\end{align}\]
As we know that
 \[\begin{align}
  & {{2}^{x}}=y \\
 & \Rightarrow {{2}^{x}}=\dfrac{1}{8},{{2}^{x}}=1 \\
 & \Rightarrow {{2}^{x}}={{\left( 2 \right)}^{-3}},{{2}^{x}}={{2}^{0}} \\
\end{align}\]
Therefore, by comparing both sides we get \[x=3,0\].

Note:
While finding the roots of the quadratic equation you can use completing the square method. After finding the value of \[y\]do not forget to substitute it in the assumption, we made to find the value of\[x\]. in order to verify the answer, but the value of \[x\] into the original equation which is given in the question.