Find an exact solution for
 
ln(6)+ln(6x2−3)=ln(151)ln6+ln6x2−3=ln151.

Let's solve the equation step by step.

The given equation is:

\[
\ln(6) + \ln(6x^2 - 3) = \ln(151)
\]

### Step 1: Use the logarithmic property

We can use the logarithmic property:

\[
\ln(a) + \ln(b) = \ln(a \cdot b)
\]

Applying this property:

\[
\ln(6) + \ln(6x^2 - 3) = \ln(6(6x^2 - 3))
\]

So the equation becomes:

\[
\ln(6(6x^2 - 3)) = \ln(151)
\]

### Step 2: Remove the logarithms

Since we have \(\ln(\text{something}) = \ln(\text{something else})\), we can drop the logarithms, which gives us the equation:

\[
6(6x^2 - 3) = 151
\]

### Step 3: Expand and simplify

Now, expand the left-hand side:

\[
6(6x^2 - 3) = 36x^2 - 18
\]

So the equation becomes:

\[
36x^2 - 18 = 151
\]

### Step 4: Solve for \(x^2\)

Add 18 to both sides:

\[
36x^2 = 151 + 18
\]
\[
36x^2 = 169
\]

Now, divide both sides by 36:

\[
x^2 = \frac{169}{36}
\]

### Step 5: Take the square root

Take the square root of both sides:

\[
x = \pm \sqrt{\frac{169}{36}} = \pm \frac{\sqrt{169}}{\sqrt{36}} = \pm \frac{13}{6}
\]

### Final Solution:

The exact solutions are:

\[
x = \frac{13}{6} \quad \text{or} \quad x = -\frac{13}{6}
\]

Do you want more details or have any questions? 

Here are 5 related questions you can explore:
1. How do you solve logarithmic equations with different bases?
2. What are the key properties of logarithms?
3. How can we apply logarithms to exponential functions?
4. Can we solve this equation graphically?
5. What are the applications of logarithmic equations in real life?

**Tip:** When solving logarithmic equations, always ensure the arguments inside the logarithms are positive.

Find an exact solution for ln(6) + ln(6x^2 − 3) = ln(151).

Learn how to find the exact solution to the logarithmic equation ln(6) + ln(6x^2 − 3) = ln(151). This solution involves key logarithmic properties and algebraic techniques, including solving quadratic equations. Suitable for students in Grades 9-12.

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