Question

27^(x-2/3)-9^(x-1)=2*3^(2x-1)-2*3^(3x-1)

Answer 1

$\displaystyle 27^{\frac{x-2}{3}}-9^{x-1}=2*3^{2x-1}-2*3^{3x-1}\\\\\frac{3^{x}}{9}}-3^{2x-2}=\frac{2*3^{2x}}{3}-\frac{2*3^{3x}}{3}\\\\\frac{3^{x}-3^{2x}}{9}=\frac{2*3^{2x}-2*3^{3x}}{3}\\\\3^{x}-3^{2x}=6*3^{2x}-6*3^{3x}\\\\3^{x}-3^{x})^{2}=6*((3^{x})^{2}-(3^{x})^{3})\\\\3^{x}(1-3^{x})=6*3^{2x}(1-3^{x})\\\\3^{x}=y\\\\y-y^{2}=6y^{2}(1-y)\\y-y^{2}-6y^{2}+6y^{3}=0\\y(6y^{2}-7y+1)=0\\\\y\neq0\\\\6y^{2}-7y+1=0\\D=b^{2}-4ac=49-24=25\\\\y_{1,2}=\frac{-bб\sqrt{D}}{2a}$

$\displaystyle y_{1}=1 \\\\ y_{2}= \frac{1}{6} \\ \\3^{x}=1 \\x_{1} =0 \\ \\3^{x}= \frac{1}{6} \\ \\x_{2}=-log_{3}6$

Ответ: {0; -log₃6}

Answer 2

To solve the equation 27^(x-2/3) - 9^(x-1) = 23^(2x-1) - 23^(3x-1), we can simplify the expressions using the properties of exponents.

First, let's simplify the terms involving powers of 9:

27^(x-2/3) can be expressed as (3^3)^(x-2/3), which simplifies to 3^(3*(x-2/3)), which further simplifies to 3^(3x-2).

Similarly, 9^(x-1) can be expressed as (3^2)^(x-1), which simplifies to 3^(2*(x-1)), which further simplifies to 3^(2x-2).

Substituting these simplified expressions back into the original equation, we have:

3^(3x-2) - 3^(2x-2) = 23^(2x-1) - 23^(3x-1).

Now, let's isolate the terms involving powers of 3 on one side of the equation:

3^(3x-2) - 3^(2x-2) - 23^(2x-1) + 23^(3x-1) = 0.

Next, we can combine like terms:

3^(3x-2) - 3^(2x-2) - 23^(2x-1) + 23^(3x-1) = 0.

Now, we have a polynomial equation in terms of 3^(x):

3^(3x-2) - 3^(2x-2) - 23^(2x-1) + 23^(3x-1) = 0.

We can solve this equation using numerical or graphical methods. Unfortunately, it doesn't have a simple algebraic solution.