To solve the equation \( 5^x + 5^5 = 5^{12} \), we will first express \( 5^5 \) in terms of \( 5^{12} \).
We know that:
\[ 5^{12} = 5^x + 5^5 \]
We can factor \( 5^5 \) out of both terms:
\[ 5^{12} = 5^5(5^{x-5} + 1) \]
Next, we can divide both sides of the equation by \( 5^5 \), given that \( 5^5 \) is not zero:
\[ \frac{5^{12}}{5^5} = 5^{x-5} + 1 \]
Using the property of exponents, we simplify:
\[ 5^{12-5} = 5^{x-5} + 1 \]
Thus, we find:
\[ 5^7 = 5^{x-5} + 1 \]
Next, we solve for \( 5^{x-5} \):
\[ 5^{x-5} = 5^7 - 1 \]
At this point, we need to simplify \( 5^7 - 1 \). We can recognize that \( 5^7 = 78125 \):
\[ 5^{x-5} = 78125 - 1 = 78124 \]
Now, we can take the logarithm to solve for \( x \):
\[ x - 5 = \log_5(78124) \]
Adding 5 to both sides gives:
\[ x = \log_5(78124) + 5 \]
For practical calculation, it might help to approximate \( \log_5(78124) \) using the change of base formula:
\[ \log_5(78124) = \frac{\log_{10}(78124)}{\log_{10}(5)} \]
Calculating the logarithm values using a calculator:
\(\log_{10}(78124) \approx 4.893\) and \(\log_{10}(5) \approx 0.699\).
Thus,
\[ \log_5(78124) \approx \frac{4.893}{0.699} \approx 7 \]
So we have:
\[ x \approx 7 + 5 = 12 \]
Therefore, the value of \( x \) that makes the original exponential equation true is:
\[ \boxed{12} \]
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