We are tasked with solving the equation:

\[
36^x - 6^x = 6
\]

### Step 1: Express \(36\) as a power of \(6\)
Since \(36 = 6^2\), we can rewrite the equation in terms of base \(6\):

\[
(6^2)^x - 6^x = 6
\]

This simplifies to:

\[
6^{2x} - 6^x = 6
\]

### Step 2: Substitution
Let \(y = 6^x\). This substitution allows us to rewrite the equation as:

\[
y^2 - y = 6
\]

### Step 3: Solve the quadratic equation
The equation becomes a quadratic in \(y\):

\[
y^2 - y - 6 = 0
\]

We can factor this quadratic:

\[
(y - 3)(y + 2) = 0
\]

### Step 4: Solve for \(y\)
From the factored form, we get two possible solutions for \(y\):

\[
y - 3 = 0 \quad \Rightarrow \quad y = 3
\]
\[
y + 2 = 0 \quad \Rightarrow \quad y = -2
\]

Since \(y = 6^x\) and exponential terms cannot be negative, we discard \(y = -2\).

Thus, we have \(y = 3\), which means:

\[
6^x = 3
\]

### Step 5: Solve for \(x\)
Now take the natural logarithm of both sides:

\[
\ln(6^x) = \ln(3)
\]

Using the logarithmic property \(\ln(a^b) = b\ln(a)\), we get:

\[
x \ln(6) = \ln(3)
\]

Solve for \(x\):

\[
x = \frac{\ln(3)}{\ln(6)} \approx 0.613
\]

### Final Answer:
The solution to the equation is approximately:

\[
x \approx 0.613
\]

Would you like a deeper explanation of any step, or have any questions?

Here are 5 related questions you can explore:

1. What is the general method for solving exponential equations?
2. How does substitution help in solving complex equations?
3. Why can't \(6^x\) equal a negative number?
4. Can you solve a similar equation, like \(4^x - 2^x = 2\)?
5. How does logarithmic calculation work for solving exponential equations?

**Tip**: In exponential equations, recognizing common bases can simplify the process!

Learn how to solve the exponential equation 36^x - 6^x = 6 using substitution and logarithms. This step-by-step guide covers key concepts such as rewriting powers, solving quadratic equations, and applying logarithmic identities. Suitable for students in Grades 10-12.

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