Question

Log0,8x+log0,8(x+1)=log0,8(x+3)

Answer 1

log0,8x+log0,8(x+1)=log0,8(x+3)

одз x + 3 > 0

x + 1 > 0

x > 0

x∈(0, +∞)

log0,8 x*(x+1)=log0,8 (x+3)

x^2 + x = x + 3

x² = 3

x12 = +- √3

-√3 < 0 не подходит

х = √3

Answer 2

To solve the equation log0.8x + log0.8(x + 1) = log0.8(x + 3), we can use the properties of logarithms.

First, we need to simplify the equation using the properties of logarithms. The sum of logarithms with the same base can be expressed as the logarithm of the product.

Using this property, we can rewrite the equation as follows:

log0.8(x(x + 1)) = log0.8(x + 3)

Now, we can remove the logarithm on both sides by raising the base (0.8) to the power of each side:

0.8^(log0.8(x(x + 1))) = 0.8^(log0.8(x + 3))

Simplifying further, the base and the logarithm cancel out, leaving us with:

x(x + 1) = x + 3

Expanding the equation:

x^2 + x = x + 3

Rearranging the terms:

x^2 = 3

Taking the square root of both sides:

x = ±√3

Therefore, the solutions to the equation log0.8x + log0.8(x + 1) = log0.8(x + 3) are x = √3 and x = -√3.