Solving the Exponential Equation: 3^{(x-1)} - 5^{(2x)} = 0 Step by Step

Exponential equations challenge our algebraic skills by combining different bases and exponents in one equation. One such problem is:

3^{(x-1)} - 5^{(2x)} = 0

Solving this type of equation requires understanding exponential growth and applying algebraic transformations effectively. In this blog, we’ll break it down into simple steps to find the exact solution.

 Step-by-Step Solution

Step 1: Rewrite the Equation

We start with:

3^{(x-1)} = 5^{(2x)}

Since the bases (3 and 5) are different, taking logarithms will help simplify the equation.

 Step 2: Apply the Natural Logarithm (ln) to Both Sides

ln(3^{(x-1)}) = ln(5^{(2x)})

Using the logarithmic rule ln(a^b) = b ln a, we simplify:

 (x-1) ln 3 = (2x) ln 5

 Step 3: Solve for x

Expanding the terms:

x ln 3 - ln 3 = 2x ln 5

Rearrange to get x on one side:

x ln 3 - 2x ln 5 = ln 3

Factor out x:

x (ln 3 - 2ln 5) = ln 3

Solve for x:

x = ln 3/{ln 3 – 2 ln3}

Using approximate logarithm values we get:

X ≈ − 0.518

 Final Answer

The solution to 3^{(x-1)} - 5^{(2x)} = 0 is:

X ≈ − 0.518

 Why This Equation is Important

Solving exponential equations like this is essential for understanding exponential growth, logarithmic properties, and algebraic transformations. These concepts appear in advanced math, physics, and real-world applications like finance and population modeling.

 Conclusion

By applying logarithmic properties and algebraic manipulation, we solved the equation step by step. Practicing similar problems will enhance your problem-solving skills and deepen your understanding of exponents and logarithms.

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