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(C) Let $f(x) = \log _2(x + 5)$ and $g(x) = 6 - x$.$f(x)$ is a strictly increasing function because the base $2 > 1$.$g(x)$ is a strictly decreasing f

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$1$

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(C) Let $f(x) = \log _2(x + 5)$ and $g(x) = 6 - x$.$f(x)$ is a strictly increasing function because the base $2 > 1$.$g(x)$ is a strictly decreasing function.Since one is strictly increasing and the other is strictly decreasing,they can intersect at most at one point.Check for integer values:If $x = 3$,$\log _2(3 + 5) = \log _2(8) = 3$ and $6 - 3 = 3$.Since $f(3) = g(3)$,$x = 3$ is a solution.Thus,there is exactly $1$ solution.

The number of solutions to the equation $\log _2(x + 5) = 6 - x$ is

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