The equation 2^x+5^x=3^x+4^x has | vedclass

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(B) Let $f(x) = 2^x + 5^x - 3^x - 4^x = 0$. We can observe that $x=0$ is a solution because $2^0 + 5^0 = 1 + 1 = 2$ and $3^0 + 4^0 = 1 + 1 = 2$. Also,

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only one non-zero real solution

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(B) Let $f(x) = 2^x + 5^x - 3^x - 4^x = 0$. We can observe that $x=0$ is a solution because $2^0 + 5^0 = 1 + 1 = 2$ and $3^0 + 4^0 = 1 + 1 = 2$. Also, $x=1$ is a solution because $2^1 + 5^1 = 2 + 5 = 7$ and $3^1 + 4^1 = 3 + 4 = 7$. To check for other solutions, consider $g(x) = 5^x - 4^x$ and $h(x) = 3^x - 2^x$. The equation is $g(x) = h(x)$. Using the Mean Value Theorem or by analyzing the derivatives, it can be shown that there are exactly two real solutions, $x=0$ and $x=1$. Since $x=0$ is a zero solution, there is only one non-zero real solution, which is $x=1$.

The equation $2^x+5^x=3^x+4^x$ has

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