Question

Log(3,(3^(x)-8))=2-x

Answer 1

3^(2-x)=3^x-8     ОДЗ 3^x-8>0     3^x>8   x>=2

9\3^x=3^x-8

3^x=a

9\a=a-8

a^2-8a-9=0

D=100

a=(8+-10)\2=(9;-1)

3^x=9

x=2

-1 не подходит по Одз

Ответ:2

Answer 2

To solve this logarithmic equation, we'll start by using the logarithmic identity:

loga(b^c) = c * loga(b)

Applying this identity to the left-hand side of the equation, we get:

log3(3^x - 8) = log3(3^(2-x))

Using the fact that loga(b) = loga(c) if and only if b = c, we can write:

3^x - 8 = 3^(2-x)

Now we have a simple equation to solve for x. First, let's simplify the right-hand side:

3^(2-x) = 3^2 / 3^x = 9 / 3^x

Substituting this into the equation above, we get:

3^x - 8 = 9 / 3^x

Multiplying both sides by 3^x, we get:

3^(2x) - 8 * 3^x - 9 = 0

Letting y = 3^x, we can rewrite this as a quadratic equation:

y^2 - 8y - 9 = 0

Factoring this quadratic equation, we get:

(y - 9)(y + 1) = 0

So y = 9 or y = -1. Since y = 3^x, we have:

3^x = 9 or 3^x = -1

The equation 3^x = -1 has no real solutions, so we can ignore it. Solving 3^x = 9, we get:

x = log3(9) = 2

Therefore, the solution to the original equation is x = 2.