Question

lg(x+√3)+lg(x-√3)=0 log₂(x-2)+log₂(x-3)=1 lg(x²-9)-lg(x-3)=0 log₆(x-1)-log₆(2x-11)=log₆2 log₀·₇log₄(x-5)=0 log²₀₋₅x-log₀₋₅x-2=0    

Answer 1

1) lg(x^2-3)=lg1

x^2-3=1

x=2

2) (x-2)(x-3)=2

x^2-5x+4=0

x1=4

x2=1 x-3<0 не подходит

3) х+3=1

х=-2 не подходит по области определения, решений нет

4) (x-1)/(2x-11)=2

x-1=4x-22

3x=21

x=7

5) log(x-5)=0,7^0=1

x-5=4^1=4

x=9

6) log=t  t^2-t-2=0

          t1=2  t2=-1 

x1=0,5^2=1/4

x2=0.5^(-1)=2

Answer 2

Let's solve each equation one by one:

1. lg(x+√3) + lg(x-√3) = 0

Using the property of logarithms, we can rewrite the equation as:

lg((x+√3)(x-√3)) = 0

Simplifying the equation inside the logarithm:

lg(x² - 3) = 0

Converting the logarithmic equation to exponential form:

10^0 = x² - 3

1 = x² - 3

x² = 4

Taking the square root of both sides:

x = ±2

So, the solutions are x = 2 and x = -2.

2. log₂(x-2) + log₂(x-3) = 1

Using the property of logarithms, we can combine the logarithms:

log₂((x-2)(x-3)) = 1

Simplifying the equation inside the logarithm:

log₂(x² - 5x + 6) = 1

Converting the logarithmic equation to exponential form:

2^1 = x² - 5x + 6

2 = x² - 5x + 6

Rearranging the equation:

x² - 5x - 4 = 0

Factoring the quadratic equation:

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

Setting each factor to zero:

x - 4 = 0 or x + 1 = 0

x = 4 or x = -1

So, the solutions are x = 4 and x = -1.

3. lg(x²-9) - lg(x-3) = 0

Using the property of logarithms, we can combine the logarithms:

lg((x²-9)/(x-3)) = 0

Simplifying the equation inside the logarithm:

lg((x+3)(x-3)/(x-3)) = 0

lg(x+3) = 0

Converting the logarithmic equation to exponential form:

10^0 = x + 3

1 = x + 3

x = -2

So, the solution is x = -2.

4. log₆(x-1) - log₆(2x-11) = log₆2

Using the property of logarithms, we can combine the logarithms:

log₆((x-1)/(2x-11)) = log₆2

Since the base of the logarithm is the same on both sides, we can equate the expressions inside the logarithms:

(x-1)/(2x-11) = 2

Solving the equation for x:

x - 1 = 4x - 22

3x = 21

x = 7

So, the solution is x = 7.

5. log₀·₇log₄(x-5) = 0

The expression log₀·₇ is undefined since the base is zero. Therefore, there is no solution to this equation.

6. log²₀₋₅x - log₀₋₅x - 2 = 0

Using the property of logarithms, we can combine the logarithms:

log₀₋₅x² - log₀₋₅x - 2 = 0

Simplifying the equation:

log₀₋₅(x²/x) - 2 = 0

log₀₋₅(x) - 2 = 0

Converting the logarithmic equation to exponential form:

(-5)^(2) = x

25 =