Solve $2^{{\displaystyle x}\large +1}-4.2^{\large -{\displaystyle x}}-7 = 0$

The two raised to the power of a variable $x$ plus one minus four times two raised to the power of the negative $x$ minus seven equals to zero should be solved to find the value of $x$ in this math problem.

It is a mathematical equation primarily but the exponential functions are involved in forming the equation. So, let us learn how to solve the exponential equation for finding the value of variable $x$.

Removal of Complexity from the equation

Look at the first term on the left hand side of the equation and the exponent is an expression. It is a sum of a variable $x$ and a number one. The number $2$ raised to the power of $x$ plus one can be split as a product of two factors by the product rule of exponents.

$\implies$ $2^{\displaystyle x} \times 2^{\large 1}$ $-$ $4 \times 2^{\large -{\displaystyle \normalsize x}}$ $-$ $7$ $\,=\,$ $0$

$\implies$ $2^{\displaystyle x} \times 2$ $-$ $4 \times 2^{\large -{\displaystyle \normalsize x}}$ $-$ $7$ $\,=\,$ $0$

$\implies$ $2 \times 2^{\displaystyle x}$ $-$ $4 \times 2^{\large -{\displaystyle \normalsize x}}$ $-$ $7$ $\,=\,$ $0$

Look at the second term. There are two factors in the second term, but the second factor is an exponential function with negative sign. So, let’s convert the exponential function with negative sign as an exponential function with positive sign by the negative power rule.

$\implies$ $2 \times 2^{\displaystyle x}$ $-$ $4 \times \dfrac{1}{2^{\displaystyle \normalsize x}}$ $-$ $7$ $\,=\,$ $0$

The $2$ raised to the power of $x$ is a factor in the first term and it is also there in the second term in reciprocal form. The reciprocal form of $2$ raised to the power of $x$ creates a problem to solve the equation. So, let us try to remove the reciprocal form from the equation and it can be done by multiplying the expressions on both sides of the equation with the $2$ raised to the power of $x$.

$\implies$ $2^{\displaystyle x} \times \bigg(2 \times 2^{\displaystyle x}$ $-$ $4 \times \dfrac{1}{2^{\displaystyle \normalsize x}}$ $-$ $7\bigg)$ $\,=\,$ $2^{\displaystyle x} \times 0$

The $2$ raised to the power of $x$ multiplies an expression on left hand side of the equation but the expression consists of three terms. So, distribute the coefficient to all the terms on the left hand side of the equation by the distributive property. On the right hand side of the equation, the $2$ raised to the power of $x$ can be multiplied by the zero directly.

$\implies$ $2^{\displaystyle x} \times 2 \times 2^{\displaystyle x}$ $-$ $2^{\displaystyle x} \times 4 \times \dfrac{1}{2^{\displaystyle \normalsize x}}$ $-$ $2^{\displaystyle x} \times 7$ $\,=\,$ $0$

According to the commutative property of multiplication, the position of each factor within the term can be changed for our convenience.

$\implies$ $2 \times 2^{\displaystyle x} \times 2^{\displaystyle x}$ $-$ $4 \times 2^{\displaystyle x} \times \dfrac{1}{2^{\displaystyle \normalsize x}}$ $-$ $7 \times 2^{\displaystyle x}$ $\,=\,$ $0$

The number $2$ raised to the power of $x$ is multiplied by itself in the first term. So, the product of them can be expressed in exponential notation. The $2$ raised to the power of $x$ is multiplied by its reciprocal in the second term. So, let us find the product of them by the multiplication of the fractions.

$\implies$ $2 \times \big(2^{\displaystyle x}\big)^2$ $-$ $4 \times \dfrac{2^{\displaystyle x} \times 1}{2^{\displaystyle \normalsize x}}$ $-$ $7 \times 2^{\displaystyle x}$ $\,=\,$ $0$

$\implies$ $2 \times \big(2^{\displaystyle x}\big)^2$ $-$ $4 \times \dfrac{\cancel{2^{\displaystyle x}}}{\cancel{2^{\displaystyle \normalsize x}}}$ $-$ $7 \times 2^{\displaystyle x}$ $\,=\,$ $0$

Now, let’s focus on simplifying the expression on the left hand side of the equation.

$\implies$ $2 \times \big(2^{\displaystyle x}\big)^2$ $-$ $4 \times 1$ $-$ $7 \times 2^{\displaystyle x}$ $\,=\,$ $0$

$\implies$ $2 \times \big(2^{\displaystyle x}\big)^2$ $-$ $4$ $-$ $7 \times 2^{\displaystyle x}$ $\,=\,$ $0$

$\,\,\,\therefore\,\,\,\,\,\,$ $2 \times \big(2^{\displaystyle x}\big)^2$ $-$ $7 \times 2^{\displaystyle x}$ $-$ $4$ $\,=\,$ $0$

Represent the exponential function by a variable

The exponential function $2$ raised to the power of $x$ is involved in the first two terms on the left hand side of the equation, and its involvement confuses us while solving the equation. So, let us try to denote the exponential function $2$ raised to the power of $x$ by a variable $y$. Now, the equation in terms of $x$ can be written as an equation in terms of a variable $y$.

$\implies$ $2 \times (y)^2$ $-$ $7 \times y$ $-$ $4$ $\,=\,$ $0$

$\implies$ $2 \times y^2$ $-$ $7 \times y$ $-$ $4$ $\,=\,$ $0$

$\,\,\,\therefore\,\,\,\,\,\,$ $2y^2$ $-$ $7y$ $-$ $4$ $\,=\,$ $0$

Solve the Quadratic equation by the factorization

The $2$ times $y$ square minus $7$ times $y$ minus $4$ equals to zero is a quadratic equation, and it can be solved by splitting the middle term to find the value of variable $y$ mathematically.

$\implies$ $2y^2$ $-$ $8y$ $+$ $y$ $-$ $4$ $\,=\,$ $0$

Now, write each term as a product of two factors in the expression on the left hand side of the equation to take the common factor out from the terms.

$\implies$ $2y \times y$ $-$ $2y \times 4$ $+$ $1 \times y$ $-$ $1 \times 4$ $\,=\,$ $0$

It is time to factorize the expression by taking out the common factors from the terms in the equation and it is useful to us in simplifying the equation further.

$\implies$ $2y \times (y-4)$ $+$ $1 \times (y-4)$ $\,=\,$ $0$

$\implies$ $(y-4) \times (2y+1)$ $\,=\,$ $0$

$\,\,\,\therefore\,\,\,\,\,\,$ $(y-4)(2y+1)$ $\,=\,$ $0$

Now, let us solve the equation to calculate the value of variable $y$.

$\implies$ $y-4$ $\,=\,$ $0$ or $2y+1$ $\,=\,$ $0$

$\implies$ $y$ $\,=\,$ $4$ or $2y$ $\,=\,$ $-1$

$\implies$ $y$ $\,=\,$ $4$ or $y$ $\,=\,$ $-$ $\dfrac{1}{2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $y$ $\,=\,$ $-$ $\dfrac{1}{2}$ or $y$ $\,=\,$ $4$

Find the solutions of the Exponential equation

The solutions of the variable $y$ are calculated by solving the quadratic equation but the given exponential equation is defined in terms of a variable $x$. So, it is time to change the variable $y$ by expressing it in terms of variable $x$.

$\implies$ $2^{\displaystyle x} \,=\, -\dfrac{1}{2}$ or $2^{\displaystyle x} \,=\, 4$

Now, we obtain two exponential equations and they are useful to us to find the value of variable $x$ by solving. Let’s solve the exponential equations one by one.

Firstly, let us focus first equation. The quantity on the left hand side of the equation is in exponential notion. So, write $1$ divided by $2$ in exponential form to solve the equation.

$\implies$ $2^{\displaystyle x} \,=\, -2^{-1}$

The bases of both sides of the equation are same but their signs are opposite. Hence, the exponents on both sides of the equation cannot be compared and it expresses that the equation in exponential form is not useful to us for solving the value of variable $x$.

Now, it is to concentrate on second equation to find the value of variable $x$ by solving the equation. Write, the number $4$ in exponential form.

$\implies$ $2^{\displaystyle x} \,=\, 2^2$

The bases of both sides of the exponential expressions are same. So, the exponents of both sides of the exponential expressions should be equal mathematically.

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, 2$

The value of a variable $x$ is equal to $2$ is the solution of the given exponential equation with quadratic form.

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