**Solving the Exponential Equation: 7***81^x + 9*49^x = 16*63^x

*81^x + 9*49^x = 16*63^x

In this article, we will explore the solution to the exponential equation:

7*81^x + 9*49^x = 16*63^x

This equation involves exponential functions with different bases and coefficients. To solve it, we will need to use some clever manipulations and properties of exponential functions.

**Step 1: Simplify the Equation**

First, let's simplify the equation by noticing that 81, 49, and 63 can be rewritten as powers of smaller numbers:

81 = 3^4 49 = 7^2 63 = 3^2 * 7

Substituting these expressions into the original equation, we get:

7*(3^4)^x + 9*(7^2)^x = 16*(3^2 * 7)^x

**Step 2: Apply Exponential Properties**

Next, we can apply the property of exponential functions that states:

(a^b)^c = a^(bc)

Using this property, we can rewrite the equation as:

7*3^(4x) + 9*7^(2x) = 16*3^(2x) * 7^x

**Step 3: Equate Coefficients**

Now, we can equate the coefficients of the terms with the same bases (3 and 7). This gives us two separate equations:

7*3^(4x) = 16*3^(2x) ... (1)
9*7^(2x) = 16*7^x ... (2)

**Step 4: Solve the Equations**

Solving equation (1), we get:

3^(4x) = (16/7)*3^(2x)

Dividing both sides by 3^(2x), we get:

3^(2x) = 16/7

Taking the logarithm base 3 of both sides, we get:

2x = log3(16/7)

x = (1/2)*log3(16/7)

Now, solving equation (2), we get:

7^(2x) = (16/9)*7^x

Dividing both sides by 7^x, we get:

7^x = 16/9

Taking the logarithm base 7 of both sides, we get:

x = log7(16/9)

**Conclusion**

We have derived two expressions for x:

x = (1/2)*log3(16/7) x = log7(16/9)

These expressions show that the value of x satisfies both equations simultaneously. Therefore, we can conclude that the given equation:

7*81^x + 9*49^x = 16*63^x

has a unique solution, which can be expressed in terms of logarithmic functions.