Question

[tex]1) 5^{x+1}+5^x=750; 2) 3^{x^{2}-22x+118}=0,(1); 3) 2^{2x+3}-15*2^{x}-2=0; [/tex]

Answer 1

$1) 5* 5^{x} +5^{x}=750$
$5^{x}(5+1)=750$
$5^{x}= \frac{750}{6} =125$
$x=3$

$2) 3^{ x^{2}-22x+118 } = 3^{-2}$
$x^{2} -22x+120=0$
$D=484-4*120= 2^{2}$
$x_{1} = \frac{22-2}{2} =10$
$x_{1} = \frac{22+2}{2} =12$

$3) 8^{2x} -15 *2^{x} -2=0$
Пусть  $2^{2x} = y^{2}$ тогда:
$8 y^{2} -15 y^{2} -2=0$
$D=225-4*-2*8= 17^{2}$
$y_{1} = \frac{15+17}{16} =2$
$y_{2} = \frac{15-17}{16} =- \frac{1}{8}$ - не подходит по условию
$2^{x} =2$
$x=1$

$4) (\frac{1}{5} )^{3-x} \ \textless \ 25$
$(\frac{1}{5} )^{3-x} \ \textless \ (\frac{1}{5} )^{-2}$
$3-x\ \textless \ -2$
$x\ \textgreater \ 5$
Промежуток от 5 до +бесконечности.

$5) x^{2} -7x+12\ \textgreater \ 0$
$D=49-4*12=1$
$x_{1} = \frac{7-1}{2} =3$
$x_{1} = \frac{7+1}{2} =4$
Чертим прямую и ставим точки.
Получится стандартный вид + - +
Промежуток от -бесконечности до 3 и от 4 до +бесконечности.

Answer 2

Let's solve each of these equations one by one:

1. $5^{x+1} + 5^x = 750$

To solve this equation, we can notice that both terms have a common base of 5. We can rewrite the equation as:

$5^x(5 + 1) = 750$

Simplify the left side:

$5^x \cdot 6 = 750$

Now, divide both sides by 6:

$5^x = \frac{750}{6} = 125$

Now, we need to express both sides with the same base (5):

$5^x = 5^3$

Since the bases are the same, we can equate the exponents:

$x = 3$

So, the solution to the equation is $x = 3$.

2. $3^{x^2 - 22x + 118} = 0$

This equation can't have a real solution because raising any real number (in this case, 3) to any power will always result in a positive value or zero, but it can never be equal to zero.

Therefore, there are no real solutions to this equation.

3. $2^{2x+3} - 15 \cdot 2^x - 2 = 0$

Let's make a substitution to simplify the equation. We can substitute $2^x$ with another variable, say $y$. This gives us:

$y^2 - 15y - 2 = 0$

Now, we can solve this quadratic equation for $y$. You can use the quadratic formula:

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

In this case, $a = 1$, $b = -15$, and $c = -2$. Plugging these values into the formula:

$y = \frac{15 \pm \sqrt{15^2 - 4(1)(-2)}}{2(1)}$

$y = \frac{15 \pm \sqrt{225 + 8}}{2}$

$y = \frac{15 \pm \sqrt{233}}{2}$

So, we have two possible values for $y$:

$y_1 = \frac{15 + \sqrt{233}}{2}$

$y_2 = \frac{15 - \sqrt{233}}{2}$

Now, we need to find the corresponding values of $x$, which were substituted as $2^x$. So:

For $y_1$:

$2^x = \frac{15 + \sqrt{233}}{2}$

For $y_2$:

$2^x = \frac{15 - \sqrt{233}}{2}$

To solve for $x$, you can take the logarithm base 2 of both sides:

For $y_1$:

$x = \log_2\left(\frac{15 + \sqrt{233}}{2}\right)$

For $y_2$:

$x = \log_2\left(\frac{15 - \sqrt{233}}{2}\right)$

These are the solutions for $x$ in the third equation. You can calculate their approximate numerical values using a calculator.