0.04^(x)−26·(0.2)^(x)+25≤0 Solve for x

Answer 1

Since 0.04^x can be re-written as (0.2^x)², our inequality can be rewritten as:

(0.2^x)² - 26(0.2^x) + 25 ≤ 0

Now, we can find where it is equal to zero. We can start by letting w=0.2^x. That gives us:

w² -26w+25 = 0

Solving for w gives us:

w = 25

w = 1

We plug w=0.2^x back in to get:

0.2^x = 25

0.2^x = 1

If we do log_0.2 on both sides, we get:

log_0.2 (0.2^x) = log_0.2(25) → x = - 2

log_0.2 (0.2^x) = log_0.2(1) → x = 0

Now we have the values where it is equal to zero, all we have to do is find which intervals of x-values yield y-values that are less than 0 when plugged into the original inequality. The three intervals we need to check are:

(-∞, -2), (-2, 0), and (0, ∞)

If we pick one number from each interval, for example: -3, -1, and 1, and plug them into the inequality, we get:

x = - 3

0.04^{- 3} - 26(0.2)^{- 3} + 25 = 12500 > 0

(since 12500 > 0, we can disregard the interval (-∞, -2) as an answer)

x = - 1

0.04^{- 1} - 26(0.2)^{- 1} + 25 = - 80 < 0

(since -80 < 0, we can include the closed interval [- 2, 0] as an answer)

x = 1

0.04¹ - 26(0.2)¹ + 25 = -19.84 > 0

(this means we can disregard the interval (0, ∞), as an answer)

So, to conclude, the only interval where the inequality is less than or equal to zero is [- 2, 0]