If f(x) = (frac{3}{5})^x + (frac{4}{5})^x - 1 | vedclass

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(B) Given $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$. Setting $f(x) = 0$,we get $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$. Dividing by $5^x$ is equi

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(B) Given $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$. Setting $f(x) = 0$,we get $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$. Dividing by $5^x$ is equivalent to solving $3^x + 4^x = 5^x$. Divide both sides by $5^x$: $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$. Let $g(x) = (\frac{3}{5})^x + (\frac{4}{5})^x$. Since both $(\frac{3}{5})^x$ and $(\frac{4}{5})^x$ are strictly decreasing functions for $x \in R$,their sum $g(x)$ is also a strictly decreasing function. $A$ strictly decreasing function can intersect the horizontal line $y = 1$ at most once. By inspection,for $x = 2$,we have $(\frac{3}{5})^2 + (\frac{4}{5})^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$. Thus,$x = 2$ is the unique solution.

If $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$,$x \in R$,then the equation $f(x) = 0$ has

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