Question

Log2 (4 - x) = 7log5 (4 + x) = 2log5 (5 - x) = log 3log2 (15+ x) = log2 3​

Answer 1

Log₂(4 - x) = 7

2⁷ = 4 - x

128 = 4 - x

x = 4 - 128

x = -124

Ответ: x = -124

Answer 2

To solve the given equations, we need to use the properties of logarithms. Let's solve each equation step by step:

1. log2(4 - x) = 7log5(4 + x)

Using the property: log(a) + log(b) = log(a * b)

log2(4 - x) = log2((4 + x)^7) (Since log5(4 + x) can be written as log2(4 + x) / log2(5) and log2(5) is a constant)

Now, if two logarithms with the same base are equal, then the arguments must be equal:

4 - x = (4 + x)^7

2. log5(4 + x) = 2log5(5 - x)

Using the property: n*log(a) = log(a^n)

log5(4 + x) = log5((5 - x)^2)

Now, if two logarithms with the same base are equal, then the arguments must be equal:

4 + x = (5 - x)^2

3. log2(15 + x) = log2(3)

Since the bases are the same and the logarithms are equal, we set the arguments equal to each other:

15 + x = 3

Now, let's solve the equations one by one:

1. 4 - x = (4 + x)^7

Expanding the right side:

4 - x = (16384 + 8192x + 1792x^2 + 224x^3 + 16x^4 + x^5)

To solve this equation, we'll have to find the roots of the 5th degree polynomial, which is a bit complicated. Solving this directly is beyond the scope of a simple solution. However, you can use numerical methods or graphing techniques to approximate the solution.

2. 4 + x = (5 - x)^2

Expanding the right side:

4 + x = (25 - 10x + x^2)

Moving all terms to one side:

x^2 + 11x + 21 = 0

Using the quadratic formula to find the roots:

x = (-11 ± √(11^2 - 4 * 1 * 21)) / 2

x = (-11 ± √(121 - 84)) / 2

x = (-11 ± √37) / 2

So, we have two potential solutions for x:

x ≈ -6.791 x ≈ -4.209

3. 15 + x = 3

Subtract 15 from both sides:

x = 3 - 15

x = -12

So, the solutions to the given equations are approximately:

x ≈ -6.791, -4.209, and -12.