Question

Log5(X)=-log0,2(14-x)

Answer 1

Решение находится во вложении

Answer 2

Основание 0,2=1/5 можно представить как 5 в степени -1 степень можно вынести перед логарифм получим:
log5(x)=log5(14-x)
логорифмы с одинаковыми основаниями равны когда подлогорифмные выражения равны, получаем:
x=14-x
2x=14
x=7

Answer 3

To solve the equation log₅(x) = -log₀.₂(14 - x), we can use the logarithmic property that states logₐ(b) = log(c) / log(a), where log(c) represents the logarithm of c to any base. Applying this property, we can rewrite the equation as follows:

log₅(x) = -log₀.₂(14 - x)

log(x) / log(5) = -log(14 - x) / log(0.₂)

Now, let's convert the base of log(0.₂) to base 10:

log(x) / log(5) = -log(14 - x) / (log(2) / log(10))

Simplifying further:

log(x) / log(5) = -log(14 - x) / (log(2) / 0.3010)

Now, we can simplify the equation by cross-multiplying:

log(x) * (log(2) / 0.3010) = -log(14 - x) * log(5)

Multiply both sides by 0.3010 to isolate the log(x):

log(x) * log(2) = -log(14 - x) * log(5) * 0.3010

Now, raise both sides as powers of 10 to eliminate the logarithms:

10^(log(x) * log(2)) = 10^(-log(14 - x) * log(5) * 0.3010)

Simplifying further:

x^log(2) = (14 - x)^(-log(5) * 0.3010)

To solve for x, we need to apply further algebraic manipulations. However, it is important to note that the given equation seems to be invalid because log₀.₂(14 - x) is undefined. The logarithm function in base 0.₂ is only defined for positive values, and since 14 - x can be negative, the equation is not solvable in its current form. Please verify the equation provided or check for any transcription errors.