$2^{\displaystyle (x+1)} -4.2^{\displaystyle -x} -7 = 0$
It is an algebraic equation, which contains exponential terms with literals. It can be solved to find the value of $x$ by transforming this equation solvable.
So, apply the product rule of exponents to express the first term in simple form.
$\implies 2^{\displaystyle x} \times 2^{\displaystyle 1} -4.2^{\displaystyle -x} -7 = 0$
Use negative power rule of exponents to express the second term in its alternative form.
$\implies 2^{\displaystyle x} \times 2^{\displaystyle 1} -4 \times \dfrac{1}{2^{\displaystyle x}} -7 = 0$
$\implies 2.2^{\displaystyle x} -4 \times \dfrac{1}{2^{\displaystyle x}} -7 = 0$
Multiply both sides of this equation by the term $2^{\displaystyle x}$ to eliminate the term $2^{\displaystyle x}$ from the denominator of the second term.
$\implies 2^{\displaystyle x} \times (2.2^{\displaystyle x} -4 \times \dfrac{1}{2^{\displaystyle x}} -7) = 0 \times 2^{\displaystyle x}$
$\implies 2.2^{\displaystyle 2x} -4 \times \dfrac{2^{\displaystyle x}}{2^{\displaystyle x}} -7.2^{\displaystyle x} = 0$
$\implies \require{cancel} 2.2^{\displaystyle 2x} -4 \times \dfrac{\cancel{2^{\displaystyle x}}}{\cancel{2^{\displaystyle x}}} -7.2^{\displaystyle x} = 0$
$\implies 2.2^{\displaystyle 2x} -4 \times 1 -7.2^{\displaystyle x} = 0$
$\implies 2.2^{\displaystyle 2x} -4 -7.2^{\displaystyle x} = 0$
$\implies 2.2^{\displaystyle 2x} -7.2^{\displaystyle x} -4 = 0$
Try power rule of exponents to express the first term in power of power of an exponential term.
$\implies 2.{(2^{\displaystyle x})}^2 -7.2^{\displaystyle x} -4 = 0$
Take $2^{\displaystyle x} = m$ and express the simplified equation in terms of $m$.
$\implies 2m^2 -7m -4 = 0$
The equation is a quadratic equation and it can be solved in any method of the quadratic equations. Here, the factors method is used to express this quadratic equation in factors and then find the value of the $x$.
$\implies 2m^{\displaystyle 2} -7m -4 = 0$
$\implies 2m^{\displaystyle 2} -8m + m -4 = 0$
$\implies 2m(m -4) + m -4 = 0$
$\implies 2m(m -4) + 1(m -4) = 0$
$\implies (2m+1)(m -4) = 0$
$\implies 2m+1 = 0$ and $m -4 = 0$
$\implies 2m = -1$ and $m = 4$
$\implies m = -\dfrac{1}{2}$ and $m = 4$
Replace the $m$ by its actual value.
$\implies 2^{\displaystyle x} = -\dfrac{1}{2}$ and $2^{\displaystyle x} = 4$
$\implies 2^{\displaystyle x} = -2^{-1}$ and $2^{\displaystyle x} = 2^{2}$
Compare both sides of the each equation. The equation $2^{\displaystyle x} = -2^{-1}$ cannot be solved mathematically. So, consider the second equation. The bases of this equation is same. Therefore, the exponents of them should be equal and the value of $x$ is $2$ from this.
Therefore, $x = 2$ is the solution of this problem.
Substitute $x = 2$ in the equation.
$\implies 2^{(2+1)} -4.2^{-2} -7 = 2^3 -\dfrac{4}{2^2} -7$
$\implies 2^{(2+1)} -4.2^{-2} -7 = 8 -\dfrac{4}{4} -7$
$\implies 2^{(2+1)} -4.2^{-2} -7 = 8 -1 -7$
$\implies 2^{(2+1)} -4.2^{-2} -7 = 8 -8$
$\therefore \,\,\,\,\, 2^{(2+1)} -4.2^{-2} -7 = 0$
Therefore, it is proved that the value of $x$ is $2$ for this problem.
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