Question

1/(3^x+5)<1/^(3^(x+1)-1)

Answer 1

Применены : замена переменной, метод интервалов, свойства степени

Answer 2

To solve the inequality $\frac{1}{3^x + 5} < \frac{1}{3^{x+1} - 1}$, you can start by multiplying both sides of the inequality by $(3^x + 5)(3^{x+1} - 1)$. Be careful with multiplying by a negative number, as it will change the direction of the inequality. Let's go step by step:

$\frac{1}{3^x + 5} < \frac{1}{3^{x+1} - 1}$

Multiply both sides by $(3^x + 5)(3^{x+1} - 1)$, remembering to reverse the inequality because you're multiplying by a negative number:

$(3^x + 5)(3^{x+1} - 1) > 1$

Now, expand and simplify:

$3^{x+1}(3^x) - 3^x + 5(3^{x+1}) - 5 > 1$

Now, you can combine like terms:

$3^{x+1} \cdot 3^x - 3^x + 5 \cdot 3^{x+1} - 5 > 1$

Now, use the properties of exponents to simplify the left side:

$3^{2x} - 3^x + 5 \cdot 3^{x+1} - 5 > 1$

Now, subtract 1 from both sides:

$3^{2x} - 3^x + 5 \cdot 3^{x+1} - 5 - 1 > 0$

$3^{2x} - 3^x + 5 \cdot 3^{x+1} - 6 > 0$

Now, factor out common terms:

$3^x(3^x - 1) + 5 \cdot 3^x(3 - 1) - 6 > 0$

Now, factor further:

$3^x(3^x - 1) + 10 \cdot 3^x - 6 > 0$

Now, you can see that you have a common factor of $3^x$, so you can simplify further:

$3^x(3^x - 1 + 10) - 6 > 0$

$3^x(3^x + 9) - 6 > 0$

Now, you have a quadratic equation in terms of $3^x$. Let's make a substitution to simplify further. Let $u = 3^x$, so the equation becomes:

$u(u + 9) - 6 > 0$

Now, solve this quadratic inequality for $u$:

$u^2 + 9u - 6 > 0$

To solve this quadratic inequality, you can find its roots by setting it equal to zero:

$u^2 + 9u - 6 = 0$

Now, use the quadratic formula to find the roots:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 1$, $b = 9$, and $c = -6$:

$u = \frac{-9 \pm \sqrt{9^2 - 4(1)(-6)}}{2(1)}$

$u = \frac{-9 \pm \sqrt{81 + 24}}{2}$