Question

a classmate of yours is having difficulty solving the following exponential equation and asks for your help in solving it. in your own words, how would you explain the process for solving the problem?
$9^{3x + 1} = 27^{3x - 1}$

Explanation:

Step1: Express bases as powers of 3

Since \(9 = 3^2\) and \(27 = 3^3\), rewrite the equation:
\((3^2)^{3x + 1} = (3^3)^{3x - 1}\)

Step2: Simplify exponents (power of a power)

Using \((a^m)^n = a^{mn}\), simplify both sides:
\(3^{2(3x + 1)} = 3^{3(3x - 1)}\)

Step3: Set exponents equal (same base, so exponents must be equal)

Since \(3^a = 3^b \implies a = b\), we get:
\(2(3x + 1) = 3(3x - 1)\)

Step4: Solve the linear equation

Expand both sides:
\(6x + 2 = 9x - 3\)

Subtract \(6x\) from both sides:
\(2 = 3x - 3\)

Add 3 to both sides:
\(5 = 3x\)

Divide by 3:
\(x = \frac{5}{3}\)

Answer:

The solution to the equation \(9^{3x + 1} = 27^{3x - 1}\) is \(x = \frac{5}{3}\). The process involves rewriting the bases as powers of 3, simplifying using exponent rules, equating the exponents, and solving the resulting linear equation.