Question

Solve \(4^{x}-4^{\sqrt{x}+1}=3 \times 2^{x+\sqrt{x}}\)

Short answer

Question: Solve the equation \(4^x - 4^{\sqrt{x}+1} = 3 \times 2^{x+\sqrt{x}}\). Answer: x = 0

Step-by-step solution

Rewrite the terms as powers of 2

Recall that \(4^x = (2^2)^x = 2^{2x}\), so we can rewrite \(4^x\) and \(4^{\sqrt{x}+1}\) in terms of powers of 2: $$2^{2x}-2^{2(\sqrt{x}+1)}=3\times 2^{x+\sqrt{x}}$$ We can now rewrite the last term \(2^{x+\sqrt{x}}\) as \(2^{x} \cdot 2^{\sqrt{x}}\)

Rewrite the equation as a quadratic equation

Let \(y = 2^x\). Since we want to rewrite the equation in terms of \(y\), we also need to express \(2^{\sqrt{x}}\) as a function of \(y\). Recall that \((2^x)^{1/2} = 2^{x/2}\). So, \(2^{\sqrt{x}} = y^{1/2}\). Now, we can rewrite the equation as: $$y^2 - y \cdot (y^{1/2})^2 = 3 \times y \cdot y^{1/2}$$ Simplifying further, we get: $$y^2 - y^2 = 3y^{3/2}$$

Solve the quadratic equation derived from the simplified equation

Since \(y^2 - y^2 = 0\), we can isolate the \(y^{3/2}\) term: $$y^{3/2} = 0$$ To solve for \(y\), we should find the square root of both sides, so we get: $$y^{3/4} = 0$$ Since any number raised to the power of 0 is equal to 1, we can write this equation as: $$y^{3/4} = 1$$ To solve for \(y\), we should find the inverse of the power, so we get: $$y = 1^{4/3}$$ Since any non-zero number raised to any power is equal to itself, we get: $$y = 1$$

Solve for x

Since we have found the value for \(y\), we can now solve for \(x\). Recall that \(y = 2^x\). We can rewrite the equation as: $$1 = 2^x$$ Taking the base 2 logarithm of both sides, we get: $$\log_2{1} = x$$ Since the logarithm of 1 to any base is always 0, we have: $$x = 0$$ The solution to the given equation is \(x = 0\).

Key concepts

Powers of 2

When dealing with exponential equations, rewriting terms as powers of a common base simplifies calculations. In the given exercise, powers of 4 are rewritten as powers of 2 because 4 is equivalent to \(2^2\). By recognizing this, large and complex exponents can be more easily managed through conversion into a common base:

• \(4^x\) rewrites as \((2^2)^x = 2^{2x}\).
• Similarly, \(4^{\sqrt{x}+1}\) is rewritten as \(2^{2(\sqrt{x}+1)}\).

By converting to a common base, the problem becomes more tractable and consistent, facilitating further simplification and solution of the equation.

Quadratic Equations

In solving exponential equations, recasting them into a quadratic form can simplify the steps—particularly when dealing with polynomial-type characteristics. Here, the method involves setting up a substitution. By letting \(y = 2^x\), the exponential equation transforms into a quadratic equation:

• This substitution leads to \(2^{\sqrt{x}} = y^{1/2}\), thus allowing the initial equation to be re-expressed in terms of \(y\): \(y^2 - y \cdot (y^{1/2})^2 = 3 \cdot y \cdot y^{1/2}\).

Simplifying further, the equation becomes \(y^2 - y^2 = 3y^{3/2}\). Quadratic equations often appear in various mathematical contexts, including those involving exponential terms, making the understanding of these equations crucial.

• By solving this, you move towards finding the values of \(y\) that satisfy the equation, which subsequently helps in determining the original variable \(x\).

Logarithms

Finally, logarithms are used to solve for the original variable in exponential equations once transformed and simplified. Here, once we deduced that \(y = 1\), we remember that \(y = 2^x\). Thus, the equation becomes \(2^x = 1\).

• Applying logarithms, specifically \(\log_2\), to both sides helps uncover the value of \(x\).

The rule that \(\log_2{1} = 0\) comes into play, since the logarithm of 1 to any base is 0. Consequently:

• This logic directly concludes that \(x = 0\), finalizing the solution to the problem.

Logarithms are a powerful tool for resolving complex exponential equations, simplifying and making clear the path to the solution.