To solve the equation \( 2^x = 5^{x+1} \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation as:
\[
2^x = 5^{x+1} = 5^x \cdot 5^1 = 5^x \cdot 5
\]

### Step 2: Divide both sides by \( 5^x \)
Now, divide both sides of the equation by \( 5^x \):
\[
\frac{2^x}{5^x} = 5
\]
\[
\left( \frac{2}{5} \right)^x = 5
\]

### Step 3: Take the natural logarithm of both sides
Take the natural logarithm (ln) on both sides:
\[
\ln\left( \left( \frac{2}{5} \right)^x \right) = \ln(5)
\]
By using the logarithm power rule \( \ln(a^b) = b \ln(a) \), this simplifies to:
\[
x \ln\left( \frac{2}{5} \right) = \ln(5)
\]

### Step 4: Solve for \( x \)
Now, isolate \( x \):
\[
x = \frac{\ln(5)}{\ln\left( \frac{5}{2} \right)}
\]

### Step 5: Approximate the value
Using a calculator to approximate the logarithms:
\[
x \approx \frac{\ln(5)}{\ln\left( \frac{5}{2} \right)} \approx \frac{1.6094}{0.9163} \approx 1.756
\]

Thus, the approximate solution is:
\[
x \approx 1.756
\]

Would you like further details on any step, or have additional questions?

Here are 5 related questions for further exploration:
1. How can we solve the equation \( 2^x = 7^{x+2} \)?
2. What happens when we change the base to logarithms in equations like \( a^x = b^{x+c} \)?
3. Can logarithms be used to solve exponential equations with complex numbers?
4. How does the value of \( x \) change if we modify the equation to \( 2^x = 5^x + 1 \)?
5. What is the graphical interpretation of solving \( 2^x = 5^{x+1} \)?

**Tip:** Always take the natural logarithm (ln) or logarithm of both sides when you have an exponential equation to simplify it.

Explore the step-by-step solution to the exponential equation 2^x = 5^{x+1}. Learn key algebraic concepts like logarithmic properties and how to simplify exponential equations using logarithms. Suitable for Grades 10-12.

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