Question

Решите уравнение: 1) 27^x=81 2) 4^x-3*2^x=4 3) log5 (x^2+11x-3)=log5 9.

Answer 1

1)3^3х=3^4

3х=4

Х=4/3

Х=1. 1/3 (1целая 1/3)

2)2^2х-3*2^х-4=0

2^х=t

t^2-3t-4=0

D=9+16=25

Подкорнем25=5

t1=(3+5)/2=8/2=4

t2=(3-5)/2=-2/2=-1неудов.

2^х=4

2^х=2^2

Х=2

3)х^2+11х-3=9

Х^2+11х-3-9=0

Х^2+11х-12=0

D=121+48=169

Подкорнем169=13

Х1=(-11+13)/2=2/2=1

Х2=(-11-13)/2=-24/2=-12

Answer 2

1. 27^x = 81

We can rewrite 27 as 3^3 and 81 as 3^4, so we have:

3^(3x) = 3^4

Equating the exponents gives:

3x = 4

So the solution is:

x = 4/3

2. 4^x - 3*2^x = 4

We can rewrite 4 as 2^2, so we have:

(2^2)^x - 3*2^x = 4

Simplifying the left side using exponent laws:

2^(2x) - 3*2^x = 4

We can substitute y = 2^x to get:

y^2 - 3y - 4 = 0

This is a quadratic equation that we can factor:

(y - 4)(y + 1) = 0

So either y - 4 = 0 or y + 1 = 0.

If y - 4 = 0, then y = 4, which means 2^x = 4. Solving for x:

2^x = 4 x = log2(4) x = 2

If y + 1 = 0, then y = -1, which means 2^x = -1, which has no real solutions.

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

3. log5(x^2 + 11x - 3) = log5(9)

Using the fact that log a = log b is equivalent to a = b, we can write:

x^2 + 11x - 3 = 9

Simplifying:

x^2 + 11x - 12 = 0

This is a quadratic equation that we can factor:

(x + 12)(x - 1) = 0

So either x + 12 = 0 or x - 1 = 0.

If x + 12 = 0, then x = -12.

If x - 1 = 0, then x = 1.

Therefore, the solutions to the original equation are x = -12 and x = 1.