Question

4.2 solve exponential equations

1. $3^{2x - 1} = 3^{x + 2}$
2. $2^{3x} = 4^{x + 2}$
3. $3^{2x - 1} = \frac{1}{9}$

hint:
$\frac{1}{9} = 3^{-2}$

1. $4^{x + 1} = 8^{2x + 3}$

hint:
$2^2 = 4$
$2^3 = 8$

1. $8^{x - 2} = \frac{1}{16}$

hint:
$\frac{1}{16} = 2^{-4}$
$8 = 2^3$

Explanation:

Problem 1: \( 3^{2x - 1} = 3^{x + 2} \)

Step 1: Use the property of exponents (if \( a^m = a^n \), then \( m = n \) for \( a > 0, a

eq 1 \))
Since the bases are the same (\( 3 \)), we can set the exponents equal to each other:
\( 2x - 1 = x + 2 \)

Step 2: Solve for \( x \)

Subtract \( x \) from both sides:
\( 2x - x - 1 = x - x + 2 \)
\( x - 1 = 2 \)

Add \( 1 \) to both sides:
\( x - 1 + 1 = 2 + 1 \)
\( x = 3 \)

Step 1: Rewrite \( 4 \) as \( 2^2 \)

\( 4^{x + 2} = (2^2)^{x + 2} \)

Step 2: Use the exponent rule \( (a^m)^n = a^{mn} \)

\( (2^2)^{x + 2} = 2^{2(x + 2)} = 2^{2x + 4} \)

Step 3: Set exponents equal (same base \( 2 \))

\( 3x = 2x + 4 \)

Step 4: Solve for \( x \)

Subtract \( 2x \) from both sides:
\( 3x - 2x = 2x - 2x + 4 \)
\( x = 4 \)

Step 1: Substitute \( \frac{1}{9} \) with \( 3^{-2} \)

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

Step 2: Set exponents equal (same base \( 3 \))

\( 2x - 1 = -2 \)

Step 3: Solve for \( x \)

Add \( 1 \) to both sides:
\( 2x - 1 + 1 = -2 + 1 \)
\( 2x = -1 \)

Divide by \( 2 \):
\( x = \frac{-1}{2} \)

Answer:

\( x = 3 \)

Problem 2: \( 2^{3x} = 4^{x + 2} \)