Let's solve the logarithmic equation
log(3) + log(x − 8) = log(x).
(a)
First, we combine the logarithms on the LHS to get the equivalent equation
= log(x).
(b)
Next, we use the fact that log is one-to-one to get the equivalent equation
= x.
(c)
Now we find x =   .
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3.
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SPRECALC8 4.5.005.

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PRACTICE ANOTHER
Find the solution of the exponential equation. (Enter your answers as a comma-separated list.)
32x − 1 = 1
x =
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SPRECALC8 4.5.008.

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PRACTICE ANOTHER
Find the solution of the exponential equation. (Enter your answers as a comma-separated list.)
e1 − 3x = e4x − 4
x =
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5.
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SPRECALC8 4.5.513.XP.

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PRACTICE ANOTHER
Consider the following.
54x − 1 = 8
(a) Find the exact solution of the exponential equation in terms of logarithms.
x =

(b) Use a calculator to find an approximation to the solution rounded to six decimal places.
x =
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### Problem 1: Logarithmic Equation  
We start with:  
\[
\log(3) + \log(x - 8) = \log(x)
\]

(a) **Combine the logarithms on the LHS**  
Using the logarithmic property \(\log(a) + \log(b) = \log(a \cdot b)\), we combine:  
\[
\log(3(x - 8)) = \log(x)
\]

(b) **Use the one-to-one property**  
Since the logarithmic function is one-to-one, we equate the arguments:  
\[
3(x - 8) = x
\]

(c) **Solve for \(x\):**  
Expanding and simplifying:  
\[
3x - 24 = x \quad \Rightarrow \quad 3x - x = 24 \quad \Rightarrow \quad 2x = 24 \quad \Rightarrow \quad x = 12
\]

**Check for extraneous solutions:**  
Substitute \(x = 12\) into the original equation:  
\[
\log(3) + \log(12 - 8) = \log(12)
\]
\[
\log(3) + \log(4) = \log(12)
\]
\[
\log(12) = \log(12) \quad \text{(True!)}
\]

Thus, the solution is:  
\[
x = 12
\]

---

### Problem 2: Exponential Equation  
We start with:  
\[
3^{2x - 1} = 1
\]

Since \(1 = 3^0\), equate the exponents:  
\[
2x - 1 = 0
\]
\[
2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}
\]

Thus, the solution is:  
\[
x = \frac{1}{2}
\]

---

### Problem 3: Exponential Equation with Base \(e\)  
We start with:  
\[
e^{1 - 3x} = e^{4x - 4}
\]

Use the one-to-one property of exponential functions to equate exponents:  
\[
1 - 3x = 4x - 4
\]

Simplify:  
\[
1 + 4 = 4x + 3x \quad \Rightarrow \quad 5 = 7x \quad \Rightarrow \quad x = \frac{5}{7}
\]

Thus, the solution is:  
\[
x = \frac{5}{7}
\]

---

### Problem 4: Exponential Equation with Logarithmic Solution  
We start with:  
\[
5^{4x - 1} = 8
\]

(a) **Solve in terms of logarithms:**  
Take the natural logarithm (\(\ln\)) on both sides:  
\[
\ln(5^{4x - 1}) = \ln(8)
\]

Using the logarithmic power rule (\(\ln(a^b) = b \ln(a)\)):  
\[
(4x - 1) \ln(5) = \ln(8)
\]

Divide through by \(\ln(5)\):  
\[
4x - 1 = \frac{\ln(8)}{\ln(5)}
\]

Solve for \(x\):  
\[
4x = \frac{\ln(8)}{\ln(5)} + 1 \quad \Rightarrow \quad x = \frac{\frac{\ln(8)}{\ln(5)} + 1}{4}
\]

(b) **Approximation:**  
Using a calculator:  
\[
\ln(8) \approx 2.0794, \quad \ln(5) \approx 1.6094
\]
\[
x \approx \frac{\frac{2.0794}{1.6094} + 1}{4} \approx \frac{1.292 + 1}{4} \approx \frac{2.292}{4} \approx 0.573
\]

Thus, the solution is:  
\[
x \approx 0.573
\]

---

### Questions for Further Practice:
1. How do you verify solutions to logarithmic equations?  
2. What is the relationship between exponential and logarithmic equations?  
3. How can extraneous solutions arise in logarithmic problems?  
4. Can you solve \(\ln(x) + \ln(x - 1) = 1\)?  
5. What are the key steps in solving \(2^{x+1} = 8\)?  

**Tip:** Always check for domain restrictions in logarithmic equations to avoid extraneous solutions!

Let's solve the logarithmic equation log(3) + log(x − 8) = log(x).

Learn how to solve logarithmic equations like log(3) + log(x − 8) = log(x) and exponential equations with detailed step-by-step solutions. Understand the properties of logarithms and exponential functions, perfect for high school students in Grades 9-12.

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