How do you solve $5^{2 x + 3} \leq 125$ $5^(2x + 3)\leq 125$ ?

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How do you solve $5^{2 x + 3} \leq 125$ $5^(2x + 3)\leq 125$ ?

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Answer 1 · 2017-03-22T02:58:36

$x \leq 0$ $x<=0$

Explanation:

We know that $5^{3} = 125$ $5^3 = 125$ . Since $5^{2 x + 3}$ $5^(2x+3)$ is less than or equal to $125$ $125$ , this must mean $2 x + 3$ $2x + 3$ is less than or equal to $3$ $3$ , as $2 x + 3 = 3$ $2x+3 = 3$ would give $125$ $125$ , and any power less than $3$ $3$ would give a number less than $125$ $125$ .

Now, we have an inequality to solve:

$2 x + 3 \leq 3$ $2x+3 <= 3$

$2 x \leq 0$ $2x <= 0$

$x \leq 0$ $x <= 0$

Therefore, we get the final answer to be $x \leq 0$ $x<=0$ .

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How do you solve $5^{2 x + 3} \leq 125$ $5^(2x + 3)\leq 125$ ?

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