Question

If \(3.2^{2 x+1}-5.2^{x+2}+16=0\) and \(x\) is an integer, find the value of \(x\). (1) 1 (2) 2 (3) 3 (4) 4

Short answer

Answer: x = 1

Step-by-step solution

Rewrite the equation in terms of \(2^x\)

First, let's rewrite the given equation: \(3.2^{2 x+1}-5.2^{x+2}+16=0\) We can rewrite \(2 x+1\) as \(x+x+1\) and \(x+2\) as \(x+1+x+1\) Now we can rewrite the equation in terms of \(2^x\) like this: \(3\cdot 2^x\cdot 2^{x+1}-5\cdot 2^{x+1}\cdot 2^{x+1}+16=0\)

Simplify the equation

Now let's simplify the equation: \(3\cdot 2^x\cdot 2^{x+1}-5\cdot 2^{x+1}\cdot 2^{x+1}+16=0\) \(3\cdot 2^x\cdot 2\times 2^{x}-5\cdot 2\times 2^{x}\times 2^{x}+16=0\)

Factor the equation

Now we can factor the common term \(2^x\): \(2^x(3\cdot 2\times 2^{x}-5\cdot 2\times 2^{x})+16=0\)

Make a substitution

To simplify the equation further, let's make a substitution: \(y = 2^x\) \(y(6y - 10y) + 16 = 0\) Now we have a simpler equation: \(-4y^2 + 16 = 0\)

Solve for y

Now we can solve for \(y\): Divide both sides by \(-4\): \(y^2 - 4 = 0\) Add 4 to both sides: \(y^2 = 4\) Take the square root of both sides: \(y = ±2\)

Substitute back and solve for x

Now we will substitute the values of \(y\) back into the equation \(y = 2^x\): For \(y = 2\): \(2^x = 2\) Taking the logarithm to base 2 of both sides: \(x = 1\) For \(y = -2\): \(2^x = -2\) A power of 2 can never be equal to a negative number, so there's no solution for this case.

Choose the correct answer

Since we found the value of \(x = 1\), the correct option is: (1) 1

Key concepts

Mathematical Exponentiation

Understanding mathematical exponentiation is crucial when solving exponential equations like the one given in the exercise. Exponentiation involves raising a number, called the base, to the power of an exponent. In our case, the base is 2, and the exponent is a function of the variable 'x'. Exponentiation follows specific rules which help in simplifying and solving equations. For example, if we have an exponent addition such as in the term \(2^{x+1}\), this is equivalent to \(2^x \cdot 2^1\). Similarly, \(2^{x+x+1}\) can be broken down into \(2^x \cdot 2^x \cdot 2^1\) or \(2^{2x+1}\). This is helpful in rewriting and simplifying the given equation to eventually solve for 'x'. Moreover, understanding that exponential functions like \(2^x\) grow very fast can aid us in estimating and checking our solutions. These concepts should be clear to effectively tackle exponential equations in the IIT Foundation Mathematics context.

Quadratic Equations

Simplification to Quadratic Form

By making a substitution such as \(y = 2^x\), the original exponential equation was transformed into a quadratic equation. A quadratic equation is in the form \(ax^2+bx+c = 0\), where 'a', 'b', and 'c' are constants. The new equation \(-4y^2 + 16 = 0\) fits this form, with \(a = -4\), \(b = 0\), and \(c = 16\).

Solving Quadratics

Solving quadratic equations is a fundamental skill in algebra. The basic methods for solving include factoring, completing the square, and using the quadratic formula. In this exercise, because the equation simplifies nicely, factoring and square root methods are the most efficient approaches. The quadratic equation \(-4y^2 + 16 = 0\) can be factored to yield \((y - 2)(y + 2) = 0\), leading to the solutions \(y = 2\) and \(y = -2\). Importantly, only positive base exponentiation equals a real number; hence, the negative solution for 'y' is not considered for the exponential equation, which reiterates why the understanding of both concepts is crucial for solving these problems.

IIT Foundation Mathematics

Students preparing for competitive exams like the IIT entrance examination need a strong foundation in mathematics. The problem discussed stems from such a foundation and exhibits the type of analytical thinking and problem-solving skills required for these exams. IIT Foundation Mathematics includes a variety of topics, such as algebra, geometry, and calculus, which often intertwine as they do in this problem. Understanding the principle of mathematical exponentiation and quadratic equations underpins the ability to solve complex problems that can be encountered in higher mathematics courses. The curriculum is designed to encourage students to think critically and to apply a range of mathematical concepts to solve problems systematically as demonstrated in solving the exponential equation. Furthermore, developing a deeper knowledge of the basics also assists in the logical progression to more advanced topics such as calculus and beyond, creating a strong base for any engineering or scientific discipline.