Question

 (1/5)^(2*x)+(1/5)^(x-2)-26<0

Answer 1

Условие: (1/5)^(2*x)-(1/5)^(x-2)-26<0
Решение: (1/5)^(2x)-((1/5)^x)/(1/5)^2-26<0
(1/5)^x=t 
⇒ t²-25t-26<0
(t-26)(t+1)<0
-1<t<26
-1<(1/5)^x<26
log(1/5)26<x
Ответ: x> log(1/5)26 , тогда только с логарифмом...

Answer 2

To solve this inequality, let's make a substitution. Let y = (1/5)^x. Then the inequality becomes:

y^2 + y^-2 - 26 < 0

Multiplying both sides by y^2, we get:

y^4 + 1 - 26y^2 < 0

Now we can solve for y^2 using the quadratic formula:

y^2 = (26 ± sqrt(676 - 4))/2 = 13 ± sqrt(165)

Since y is positive (because it's a power of 1/5), we have:

y^2 < 13 + sqrt(165)

or

y^2 > 13 - sqrt(165)

Taking the square root of both sides, we get:

(1/5)^x < sqrt(13 + sqrt(165)) or (1/5)^x > sqrt(13 - sqrt(165))

Finally, taking the logarithm base 1/5 of both sides, we get:

x < log(1/5)(sqrt(13 + sqrt(165))) or x > log(1/5)(sqrt(13 - sqrt(165)))

Using a calculator, we can approximate these values as:

x < 4.573 or x > 9.427

Therefore, the solution to the inequality is:

x < 4.573 or x > 9.427.