Problem 128 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ 5^{1+x^{3}}-5^{1-x^{3}}=24 $$

Step-by-Step Solution

Answer

The value of x that satisfies the equation \(5^{1+x^{3}}-5^{1-x^{3}}=24\) is approximately \(x \approx 0.798\).

Step 1: Rewrite the equation as a sum of exponents

The given equation is \(5^{1+x^{3}}-5^{1-x^{3}}=24\). To solve for x, let's first rewrite the equation as a sum of exponents: \[ 5^{1 - x^3} + 24 = 5^{1 + x^3} \]

Step 2: Introduce a change of variable

Next, we'll make a change of variable to simplify our equation. Let's define a new variable, y, such that: \[ y = x^3 \] So now our equation is: \[ 5^{1 - y} + 24 = 5^{1 + y} \]

Step 3: Apply logarithm properties to solve for y

To solve for y, we'll use a logarithm property which states that: \(a^{b} = c\) can be transformed into \(b = \log_{a}{c}\). Applying this property to both sides of our equation, we get: \[ \log_{5}{(5^{1 - y})} = \log_{5}{(5^{1 + y} - 24)} \] As log identities state that \(\log_{a}{a^b} = b\), we can simplify the equation further: \[ 1 - y = \log_{5}{(5^{1 + y} - 24)} \] Now we isolate y on one side of the equation: \[ y = 1 - \log_{5}{(5^{1 + y} - 24)} \]

Step 4: Solve the equation numerically

At this point, we have an equation in a single variable y, but it is not an equation that can be solved algebraically. So we will resort to a numerical solution. Using a numerical solver or graphing calculator, find the value of y that makes the equation above true: \[ y \approx 0.497 \]

Step 5: Convert back to x

Now that we have our value for y, let's convert it back to x using the relationship we established in step 2: \[ y = x^3 \Rightarrow x \approx \sqrt[3]{0.497} \] So our solution is: \[ x \approx \sqrt[3]{0.497} \approx 0.798 \] So, the value of x that satisfies the given equation is approximately 0.798.

Key Concepts

Logarithm Properties

Understanding logarithm properties can be incredibly helpful when solving exponential equations. A logarithm, written as \( \log_b{a} \) where \( b \) is the base, \( a \) is the value, essentially answers the question: 'To what power must we raise \( b \) to get \( a \)?' One of the most fundamental properties is that \( \log_{b}{b^y} = y \) because the base \( b \) raised to the power of \( y \) will give us \( b^y \) by definition.

Other important properties include the logarithm of a product, \( \log_b{xy} = \log_b{x} + \log_b{y} \) and the logarithm of a quotient, \( \log_b{\frac{x}{y}} = \log_b{x} - \log_b{y} \) which enable us to simplify complex expressions. In our equation, we can exploit the property \( \log_b{b^y} = y \) to isolate the variable and move towards a solution. By doing so, we transform an unwieldy exponential equation into a more manageable form that can be tackled with further algebra or numerical methods.

Change of Variable

A change of variable is a powerful strategy in algebra that simplifies complex equations. This technique involves substituting a part of an equation with a new variable to make the equation less complicated and more approachable. For instance, in our original problem, we replaced \( x^3 \) with a single variable \( y \) to transform a difficult exponential equation into a simpler one.

By implementing a change of variable, we can focus on solving for \( y \) without worrying about the complexity of \( x^3 \) initially. This method is practical when the new variable isolates a particular part of the equation that would otherwise be cumbersome to manipulate. After solving for the new variable, as we did with \( y \) in our example, we can reverse the substitution to find the value of the original variable, \( x \) in this case.

Numerical Solution

There are instances in mathematics where an exact algebraic solution is either incredibly complex or impossible to determine. In such cases, a numerical solution is a practical alternative. Numerical solutions involve using methods to approximate the value of a variable. These methods can include iterative processes, numerical analysis, or graphical approaches, often assisted by calculators or computers.

In our example, after manipulating the equation using logarithm properties and a change of variable, we reached a point where the solution for \( y \) couldn’t be easily expressed in a closed algebraic form. Here, a numerical solver helps us approximate the value of \( y \) which we then use to find \( x \) by reverting our change of variable. Numerical solutions provide an essential tool for solving real-world problems where precision can be balanced against practicality, and an approximate value can be sufficiently accurate for the purpose at hand.