solve log5 (3x-7)= 2 + log5 2xsolve log5 (3x-7)= 2 + log5 2x

We are tasked with solving the equation:

\[
\log_5 (3x - 7) = 2 + \log_5 (2x)
\]

### Step 1: Apply logarithmic properties
We can start by applying the **logarithmic property**: 

\[
\log_b A - \log_b B = \log_b \left( \frac{A}{B} \right)
\]

First, subtract \(\log_5 (2x)\) from both sides:

\[
\log_5 (3x - 7) - \log_5 (2x) = 2
\]

Now, using the logarithmic property, the left side becomes:

\[
\log_5 \left( \frac{3x - 7}{2x} \right) = 2
\]

### Step 2: Rewrite in exponential form
The next step is to rewrite the logarithmic equation in its exponential form. Remember that:

\[
\log_b (A) = C \quad \text{means} \quad A = b^C
\]

Thus, we can rewrite:

\[
\frac{3x - 7}{2x} = 5^2
\]

Since \(5^2 = 25\), we get:

\[
\frac{3x - 7}{2x} = 25
\]

### Step 3: Solve for \(x\)
Now, solve this equation for \(x\). First, multiply both sides by \(2x\) to eliminate the denominator:

\[
3x - 7 = 25 \cdot 2x
\]

\[
3x - 7 = 50x
\]

Now, move all terms involving \(x\) to one side:

\[
3x - 50x = 7
\]

\[
-47x = 7
\]

Finally, divide by \(-47\):

\[
x = \frac{7}{-47}
\]

Thus, the solution is:

\[
x = -\frac{7}{47}
\]

### Conclusion:
The value of \(x\) is \(x = -\frac{7}{47}\).

Would you like more details or have any questions?

Here are 5 related questions to expand on this:
1. How would the solution change if the base of the logarithms was different?
2. Can you solve a similar equation using natural logarithms (\(\ln\))?
3. What are the general properties of logarithmic functions?
4. How does the change of base formula work with logarithms?
5. How can you verify if the solution is valid for the original equation?

**Tip:** Always check if the solution is within the domain of the logarithmic functions to ensure it is valid!

Discover how to solve the logarithmic equation log5 (3x-7) = 2 + log5 (2x) step by step. This solution involves logarithmic properties and exponential rewriting, ideal for high school students in Grades 10-12.

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