To solve the equation
\[ \log_2 x + \log_4 x = \log_{10} 128, \]
we can start by rewriting \(\log_4 x\) in terms of a base of 2. Recall that:
\[ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}, \]
because \(4 = 2^2\). This allows us to rewrite the left side of the equation as follows:
\[ \log_2 x + \frac{\log_2 x}{2} = \log_2 x + 0.5 \log_2 x = \frac{3}{2} \log_2 x. \]
Now, substituting this back into the equation, we have:
\[ \frac{3}{2} \log_2 x = \log_{10} 128. \]
Next, we need to simplify \(\log_{10} 128\). Since \(128 = 2^7\), we can rewrite the logarithm as:
\[ \log_{10} 128 = \log_{10}(2^7) = 7 \log_{10} 2. \]
Now our equation looks like:
\[ \frac{3}{2} \log_2 x = 7 \log_{10} 2. \]
To solve for \(\log_2 x\), we'll multiply both sides by \(\frac{2}{3}\):
\[ \log_2 x = \frac{2 \cdot 7 \log_{10} 2}{3} = \frac{14 \log_{10} 2}{3}. \]
Now we can convert \(\log_{10} 2\) to \(\log_2\) using the change of base formula:
\[ \log_2 x = \frac{14 \cdot \frac{1}{\log_2 10}}{3}. \]
Thus:
\[ \log_2 x = \frac{14}{3 \log_2 10}. \]
To find \(x\), we raise 2 to the power on both sides:
\[ x = 2^{\frac{14}{3 \log_2 10}}. \]
Using the property \(a^{\log_b c} = c^{\log_b a}\), we have:
\[ x = 10^{\frac{14}{3}} \]
Thus, the exact answer for \(x\) is:
\[ \boxed{10^{\frac{14}{3}}}. \]
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