Question

Очень срочно!!!Помогите решить уравнения,пожалуйста!!! 1)2^-x=128 2)log2(x-1)=3 3)2*(3^x+3) + 7*(3^x-2)=493

Answer 1

$2^{-x} =2^7$
$x=-7$

$log_{2} (x-1)=log_{2} 2^3$
$log_{2} (x-1)=log_{2} 8$
ОДЗ x>1
$x-1=8$
$x=9$

$2* 3^{x+3} +7* 3^{x-2} =493$
$2*3^3* 3^{x} + 7* 3^{-2} * 3^{x} =493$
$2*3^3* 3^{x} + \frac{7}{9} * 3^{x} =493$
$54* 3^{x} + \frac{7}{9} * 3^{x} =493$
$\frac{486*3^x+7*3^x}{9} =493$
$\frac{3^x(486+7)}{9} =493$
$\frac{3^x*493}{9} =493$
$\frac{3^x}{9} =1$
$3^x=9$
$x=2$

Answer 2

Смотреть во вложении

Answer 3

1) 2^-x = 128

To solve this equation, we need to isolate the variable x. Let's start by rewriting 128 as a power of 2:

128 = 2^7

So the equation becomes:

2^-x = 2^7

To solve for x, we can equate the exponents:

-x = 7

Now, let's solve for x by multiplying both sides of the equation by -1:

x = -7

Therefore, the solution to the equation 2^-x = 128 is x = -7.

2) log2(x-1) = 3

To solve this equation, we need to eliminate the logarithm. We can do this by exponentiating both sides of the equation with base 2:

2^(log2(x-1)) = 2^3

This simplifies to:

x - 1 = 8

Now, let's solve for x by adding 1 to both sides of the equation:

x = 9

Therefore, the solution to the equation log2(x-1) = 3 is x = 9.

3) 2 * (3^x + 3) + 7 * (3^x - 2) = 493

To solve this equation, we can