Question

(1/5)^(2х+1/1-х) > 125

Answer 1

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Answer 2

To solve the inequality $(1/5)^{2x + 1/(1 - x)} > 125$, we need to find the range of values for $x$ that satisfy the inequality.

Step 1: Simplify the inequality.

Let's first simplify the left-hand side of the inequality:

$(1/5)^{2x + 1/(1 - x)}$

To combine the exponents, we'll find a common denominator for $2x$ and $1/(1 - x)$:

$\frac{1}{1 - x} = \frac{(1 - x)}{(1 - x)} = \frac{1 - x}{1 - x} = 1$

So the expression becomes:

$(1/5)^{2x + 1} = (1/5)^{2x} \times (1/5)^1 = (1/5)^{2x} \times 1/5$

Now, we can rewrite $1/5$ as $5^{-1}$:

$(5^{-1})^{2x} \times 5^{-1} = 5^{-2x} \times 5^{-1} = 5^{-2x - 1}$

Step 2: Rewrite 125 as $5^3$ because $125 = 5^3$.

Now the inequality becomes:

$5^{-2x - 1} > 5^3$

Step 3: Compare the exponents.

Since the base is the same (5), we can compare the exponents:

$-2x - 1 > 3$

Step 4: Solve for $x$.

Add 1 to both sides of the inequality:

$-2x > 3 + 1$

$-2x > 4$

Divide by -2, remembering to reverse the inequality when dividing by a negative number:

$x < \frac{4}{-2}$

$x < -2$

Thus, the solution for the inequality is $x < -2$.