Answer :
Let's solve the equation step-by-step.
The given equation is:
[tex]\[ x^{\log_{10} \left( \frac{5x}{2} \right)} = 10^{\log_{10} x} \][/tex]
To solve this, we can use properties of logarithms and exponents.
1. Simplify the right-hand side:
Since [tex]\(10^{\log_{10} x} = x\)[/tex], we can rewrite the equation as:
[tex]\[ x^{\log_{10} \left( \frac{5x}{2} \right)} = x \][/tex]
2. Assuming [tex]\(x \neq 0\)[/tex], take the logarithm base 10 of both sides to simplify:
[tex]\[ \log_{10} \left( x^{\log_{10} \left( \frac{5x}{2} \right)} \right) = \log_{10} (x) \][/tex]
3. Apply the power rule of logarithms ([tex]\(\log_{10}(a^b) = b \log_{10}(a)\)[/tex]):
[tex]\[ \log_{10} \left( \frac{5x}{2} \right) \cdot \log_{10}(x) = \log_{10}(x) \][/tex]
4. Since [tex]\( \log_{10}(x) \)[/tex] is not zero (otherwise [tex]\(x\)[/tex] would be 1, which we will check later), we can cancel [tex]\( \log_{10}(x) \)[/tex] from both sides:
[tex]\[ \log_{10} \left( \frac{5x}{2} \right) = 1 \][/tex]
5. Solve for [tex]\(\frac{5x}{2}\)[/tex]:
Recall that if [tex]\(\log_{10}(a) = b\)[/tex], then [tex]\(a = 10^b\)[/tex]. So,
[tex]\[ \frac{5x}{2} = 10 \][/tex]
6. Solve for [tex]\(x\)[/tex]:
[tex]\[ 5x = 20 \][/tex]
[tex]\[ x = 4 \][/tex]
Now, let's verify if there are any other potential solutions or contradictions. Let's check if [tex]\(x = 1\)[/tex] could be a solution by plugging it back into the original equation:
For [tex]\(x = 1\)[/tex]:
[tex]\[ 1^{\log_{10} \left( \frac{5 \cdot 1}{2} \right)} \neq 10^{\log_{10} 1} \][/tex]
This does not hold since [tex]\(\log_{10}1 = 0\)[/tex] therefore [tex]\(10^0 = 1\)[/tex]. Also, the left side would be [tex]\(1^{\log_{10}\left(2.5\right)} \neq 1\)[/tex].
No other [tex]\(x\)[/tex] satisfies the equation other than [tex]\(x = 4\)[/tex]. As [tex]\(4\)[/tex] is the only root, its sum is simply [tex]\(4\)[/tex].
Thus, the sum of the real roots of the equation is:
[tex]\[ \boxed{0} \][/tex]
The given equation is:
[tex]\[ x^{\log_{10} \left( \frac{5x}{2} \right)} = 10^{\log_{10} x} \][/tex]
To solve this, we can use properties of logarithms and exponents.
1. Simplify the right-hand side:
Since [tex]\(10^{\log_{10} x} = x\)[/tex], we can rewrite the equation as:
[tex]\[ x^{\log_{10} \left( \frac{5x}{2} \right)} = x \][/tex]
2. Assuming [tex]\(x \neq 0\)[/tex], take the logarithm base 10 of both sides to simplify:
[tex]\[ \log_{10} \left( x^{\log_{10} \left( \frac{5x}{2} \right)} \right) = \log_{10} (x) \][/tex]
3. Apply the power rule of logarithms ([tex]\(\log_{10}(a^b) = b \log_{10}(a)\)[/tex]):
[tex]\[ \log_{10} \left( \frac{5x}{2} \right) \cdot \log_{10}(x) = \log_{10}(x) \][/tex]
4. Since [tex]\( \log_{10}(x) \)[/tex] is not zero (otherwise [tex]\(x\)[/tex] would be 1, which we will check later), we can cancel [tex]\( \log_{10}(x) \)[/tex] from both sides:
[tex]\[ \log_{10} \left( \frac{5x}{2} \right) = 1 \][/tex]
5. Solve for [tex]\(\frac{5x}{2}\)[/tex]:
Recall that if [tex]\(\log_{10}(a) = b\)[/tex], then [tex]\(a = 10^b\)[/tex]. So,
[tex]\[ \frac{5x}{2} = 10 \][/tex]
6. Solve for [tex]\(x\)[/tex]:
[tex]\[ 5x = 20 \][/tex]
[tex]\[ x = 4 \][/tex]
Now, let's verify if there are any other potential solutions or contradictions. Let's check if [tex]\(x = 1\)[/tex] could be a solution by plugging it back into the original equation:
For [tex]\(x = 1\)[/tex]:
[tex]\[ 1^{\log_{10} \left( \frac{5 \cdot 1}{2} \right)} \neq 10^{\log_{10} 1} \][/tex]
This does not hold since [tex]\(\log_{10}1 = 0\)[/tex] therefore [tex]\(10^0 = 1\)[/tex]. Also, the left side would be [tex]\(1^{\log_{10}\left(2.5\right)} \neq 1\)[/tex].
No other [tex]\(x\)[/tex] satisfies the equation other than [tex]\(x = 4\)[/tex]. As [tex]\(4\)[/tex] is the only root, its sum is simply [tex]\(4\)[/tex].
Thus, the sum of the real roots of the equation is:
[tex]\[ \boxed{0} \][/tex]
