Let's solve the equation $10^{x+3} = 5e^{7-x}$.

Step 1: Take the natural logarithm (ln) of both sides.

Taking the natural logarithm of both sides gives us:

$\ln(10^{x+3}) = \ln(5e^{7-x})$

Step 2: Apply the logarithmic properties.

Using the logarithmic property $\ln(a^b) = b\ln(a)$, we simplify both sides:

$(x+3)\ln(10) = \ln(5) + (7-x)\ln(e)$

Note that $\ln(e) = 1$, so the equation simplifies to:

$(x+3)\ln(10) = \ln(5) + (7-x)$

Step 3: Expand and simplify.

Expand both sides:

$x\ln(10) + 3\ln(10) = \ln(5) + 7 - x$

Step 4: Collect like terms.

Now, collect terms with $x$ on one side and constants on the other side:

$x\ln(10) + x = 7 - 3\ln(10) + \ln(5)$

Factor out $x$ on the left-hand side:

$x(\ln(10) + 1) = 7 - 3\ln(10) + \ln(5)$

Step 5: Solve for $x$.

Finally, divide both sides by $\ln(10) + 1$:

$x = \frac{7 - 3\ln(10) + \ln(5)}{\ln(10) + 1}$

This is the exact solution. You can approximate it using the values of the logarithms:

• $\ln(10) \approx 2.3026$
• $\ln(5) \approx 1.6094$

Substituting these values into the equation:

$x \approx \frac{7 - 3(2.3026) + 1.6094}{2.3026 + 1}$

$x \approx \frac{7 - 6.9078 + 1.6094}{3.3026}$

$x \approx \frac{1.7016}{3.3026} \approx 0.515$

Thus, $x \approx 0.515$.

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Would you like a detailed explanation of any step, or do you have any other questions?

Here are 5 related questions to explore further:

1. How do you solve equations involving both logarithmic and exponential terms?
2. What is the difference between natural logarithms and logarithms to other bases?
3. How can you approximate the value of logarithms without a calculator?
4. Can you use logarithms to solve quadratic equations?
5. What are some common applications of logarithms in real-world scenarios?

Tip: Always use properties of logarithms carefully, especially when dealing with exponential equations.