Question

1. logx( х3 – 5х + 10 ) = 3 3. lg( x + 5) – lg( x2 – 25 ) = 0. ​

Answer 1

Ответ:

Пошаговое объяснение:

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ln(x³+5x-10) = 3lnx

ln(x³+5x-10) = lnx³

x³+5x-10 = x³   ⇒  x = 2

2) lg( x + 5) – lg( x² – 25 ) = 0

$\frac{ln(x+5)}{ln10} - \frac{ln(x^2-25)}{ln10} =0$

ln(x+5)-ln(x²-25)=0  ⇒   x+5 = x²-25

x² -x -30 =0  ⇒  x₁ =6;  x₂ = -5

x₂ = -5 не подходит, т.к. по определению логарифма х ≠ ±5

наш ответ  х = 6

 

Answer 2

To solve these logarithmic equations, we'll work on each of them separately:

1. log_x(x^3 - 5x + 10) = 3

First, we need to apply the properties of logarithms to isolate the expression inside the logarithm:

log_x(x^3 - 5x + 10) = 3

Using the property that log_a(b) = c is equivalent to a^c = b:

x^3 - 5x + 10 = x^3

Now, subtract x^3 from both sides:

-5x + 10 = 0

Subtract 10 from both sides:

-5x = -10

Now, divide by -5 to isolate x:

x = -10 / -5

x = 2

So, the solution to the equation is x = 2.

2. lg(x + 5) - lg(x^2 - 25) = 0

First, apply the properties of logarithms to simplify the equation:

lg(x + 5) - lg(x^2 - 25) = 0

Using the property that log_a(b) - log_a(c) = log_a(b/c):

lg((x + 5) / (x^2 - 25)) = 0

Now, we can rewrite this equation as an exponential equation:

(x + 5) / (x^2 - 25) = 10^0

Since 10^0 is equal to 1:

(x + 5) / (x^2 - 25) = 1

Now, we can cross-multiply:

x + 5 = x^2 - 25

Next, subtract x from both sides:

5 = x^2 - x - 25

Now, add 25 to both sides:

x^2 - x + 30 = 0

Now, factor the quadratic equation:

(x - 6)(x + 5) = 0

Set each factor equal to zero and solve for x:

1. x - 6 = 0 x = 6

2. x + 5 = 0 x = -5

So, there are two potential solutions: x = 6 and x = -5. However, we need to check if they satisfy the original equation since logarithms are only defined for positive arguments.

For the second equation, x + 5 must be positive:

1. For x = 6: x + 5 = 6 + 5 = 11 (positive)

2. For x = -5: x + 5 = -5 + 5 = 0 (not positive)

Therefore, the only valid solution is x = 6.