Question

How do you solve ${{2}^{x}}-{{2}^{-x}}=5$?

Answer 1

Hint: To solve the above equation, we are going to assume ${{2}^{x}}=t$ and then rewrite the above equation. After letting ${{2}^{x}}=t$ in the above equation and simplifying further we will get a quadratic equation in t. Then we will solve the quadratic equation and hence will get the two different values of “t”. Then equate these two values to ${{2}^{x}}$ and then take logarithm with base 2 on both these sides and hence will get the value of x.

Complete step by step solution:
In the above problem, we have given the following equation in x:
${{2}^{x}}-{{2}^{-x}}=5$
Let us assume ${{2}^{x}}=t$ in the above equation and then the above equation will look as follows:
$\Rightarrow t-{{t}^{-1}}=5$
In the above equation, you can see that one of the power of t is -1 and we know that if a number has power -1 then we can write the reciprocal of the number so using this property in the above equation we get,
$\Rightarrow t-\dfrac{1}{t}=5$
Taking “t” as L.C.M on the L.H.S of the above equation we get,
$\Rightarrow \dfrac{{{t}^{2}}-1}{t}=5$
On cross-multiplying the above equation we get,
$\Rightarrow {{t}^{2}}-1=5t$
Subtracting 5t on both the sides of the above equation we get,
$\begin{align}
  & \Rightarrow {{t}^{2}}-1-5t=0 \\
 & \Rightarrow {{t}^{2}}-5t-1=0 \\
\end{align}$
To solve the above quadratic equation we are going to use the Shreedharacharya rule of quadratic equation $a{{x}^{2}}+bx+c=0$ in which x is equal to:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, finding the values of “t”, we are going to substitute x as “t”, “b” as -5, “a” as 1 and “c” as -1 in the above equation and we get,
$\begin{align}
  & \Rightarrow t=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow t=\dfrac{5\pm \sqrt{25+4}}{2} \\
 & \Rightarrow t=\dfrac{5\pm \sqrt{29}}{2} \\
\end{align}$
From the above we got two values of t which are:
$t=\dfrac{5+\sqrt{29}}{2},\dfrac{5-\sqrt{29}}{2}$
Now, we are going to equate these two values of t to ${{2}^{x}}$ we get,
$\begin{align}
  & \dfrac{5+\sqrt{29}}{2}={{2}^{x}}; \\
 & \dfrac{5-\sqrt{29}}{2}={{2}^{x}} \\
\end{align}$
Solving the above two equations one by one we get,
$\Rightarrow \dfrac{5+\sqrt{29}}{2}={{2}^{x}}$
Cross multiplying the above equation will give:
$\begin{align}
  & \Rightarrow 5+\sqrt{29}=2\left( {{2}^{x}} \right) \\
 & \Rightarrow 5+\sqrt{29}={{2}^{x+1}} \\
\end{align}$
Now, taking logarithm to the base 2 on both the sides we get,
${{\log }_{2}}\left( 5+\sqrt{29} \right)={{\log }_{2}}{{\left( 2 \right)}^{x+1}}$
We know the property of the logarithm which states that:
${{\log }_{a}}{{a}^{t}}=t$
Using the above property in the logarithmic equation we get,
$\begin{align}
  & \Rightarrow {{\log }_{2}}\left( 5+\sqrt{29} \right)={{\log }_{2}}{{\left( 2 \right)}^{x+1}} \\
 & \Rightarrow {{\log }_{2}}\left( 5+\sqrt{29} \right)=x+1 \\
 & \Rightarrow {{\log }_{2}}\left( 5+5.38 \right)=x+1 \\
 & \Rightarrow {{\log }_{2}}10.38=x+1 \\
 & \Rightarrow 3.375=x+1 \\
 & \Rightarrow x=2.375 \\
\end{align}$
Now, solving the other solution of t we get,
$\Rightarrow \dfrac{5-\sqrt{29}}{2}={{2}^{x}}$
Cross multiplying the above equation we get,
$\Rightarrow 5-\sqrt{29}={{2}^{x+1}}$
Taking logarithm to the base 2 on both the sides we get,
$\begin{align}
  & \Rightarrow {{\log }_{2}}\left( 5-\sqrt{29} \right)={{2}^{x+1}} \\
 & \Rightarrow {{\log }_{2}}\left( 5-5.38 \right)={{2}^{x+1}} \\
 & \Rightarrow {{\log }_{2}}\left( -0.38 \right)={{2}^{x+1}} \\
\end{align}$
We know that the domain of logarithm cannot be negative so the above solution for x cannot be possible.
Hence, we have solved the above equation and the value of x is 2.375.

Note: The mistake which is a possibility in the above problem is that if you have given the multiple choice problem and in the options you just have to mark whether one, two or 0 solutions are there. Then after getting two values of t we just mark the option which contains 2 solutions. But this is the wrong option because we are getting only 1 solution as the answer. So, make sure you will first check all the solutions and by equating the values of t with ${{2}^{x}}$ and then mark the answer.