We are tasked with solving the equation:

\[
4^{2x - 1} = 5^{x + 2}
\]

### Step 1: Rewrite the equation
First, we rewrite \(4^{2x - 1}\) as \((2^2)^{2x - 1}\), which simplifies to:

\[
(2^2)^{2x - 1} = 2^{4x - 2}
\]

So the equation becomes:

\[
2^{4x - 2} = 5^{x + 2}
\]

### Step 2: Take the logarithm of both sides
To solve for \(x\), we take the logarithm (base 10 or natural log) of both sides. Using natural logarithm (ln):

\[
\ln(2^{4x - 2}) = \ln(5^{x + 2})
\]

### Step 3: Use the logarithmic property
Apply the logarithmic property \( \ln(a^b) = b \ln(a) \):

\[
(4x - 2) \ln(2) = (x + 2) \ln(5)
\]

### Step 4: Expand and collect terms
Now, expand both sides:

\[
(4x - 2) \ln(2) = x \ln(5) + 2 \ln(5)
\]

Distribute:

\[
4x \ln(2) - 2 \ln(2) = x \ln(5) + 2 \ln(5)
\]

### Step 5: Collect \(x\)-terms on one side
Move all terms involving \(x\) to one side and constants to the other:

\[
4x \ln(2) - x \ln(5) = 2 \ln(5) + 2 \ln(2)
\]

Factor out \(x\) on the left-hand side:

\[
x(4 \ln(2) - \ln(5)) = 2 \ln(5) + 2 \ln(2)
\]

### Step 6: Solve for \(x\)
Now, solve for \(x\) by dividing both sides:

\[
x = \frac{2 \ln(5) + 2 \ln(2)}{4 \ln(2) - \ln(5)}
\]

This expression gives the exact solution for \(x\). You can approximate the value using logarithms:

\[
\ln(2) \approx 0.6931, \quad \ln(5) \approx 1.6094
\]

Substitute these values:

\[
x \approx \frac{2(1.6094) + 2(0.6931)}{4(0.6931) - 1.6094}
\]

Simplifying:

\[
x \approx \frac{3.2188 + 1.3862}{2.7724 - 1.6094} = \frac{4.605}{1.163} \approx 3.96
\]

Thus, the approximate solution is:

\[
x \approx 3.96
\]

Would you like further details or clarification?

### Related Questions:
1. How do you solve logarithmic equations more generally?
2. What are the properties of logarithms used in solving exponential equations?
3. How would the approach change if the bases were the same on both sides of the equation?
4. Can logarithms be used to solve any exponential equation, or are there exceptions?
5. How can we simplify logarithmic expressions more efficiently?

### Tip: 
Always remember that for exponential equations, logarithms are your key tool to transform and solve them when you cannot make the bases identical.

Explore how to solve the exponential equation 4^(2x-1) = 5^(x+2) using logarithms. This step-by-step guide covers key concepts in logarithmic and exponential functions, providing a clear solution for students in Grades 10-12. Includes a breakdown of logarithmic properties and exponent rules.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation