The number of solutions of log_{2}(x + 5) = 6 | vedclass

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(D) Given the equation $\log_{2}(x + 5) = 6 - x$.Let $f(x) = \log_{2}(x + 5)$ and $g(x) = 6 - x$.$f(x)$ is a strictly increasing function for $x > -5$

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$1$

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(D) Given the equation $\log_{2}(x + 5) = 6 - x$.Let $f(x) = \log_{2}(x + 5)$ and $g(x) = 6 - x$.$f(x)$ is a strictly increasing function for $x > -5$.$g(x)$ is a strictly decreasing function.At $x = 3$,$f(3) = \log_{2}(3 + 5) = \log_{2}(8) = 3$ and $g(3) = 6 - 3 = 3$.Since $f(x)$ is strictly increasing and $g(x)$ is strictly decreasing,they intersect at exactly one point.Therefore,the number of solutions is $1$.

The number of solutions of $\log_{2}(x + 5) = 6 - x$ is

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