Question

Log[2](x+1) + log[4](1+2x) = 3

Answer 1

ОДЗ

x>-0.5

log(2)(x+1)+0.5*log(1+2x)=log(2)8

log(2)((x+1)*√(1+2x))=log(2)8

(x+1)√(1+2x)=8

все в квадрат

(x+1)^2(1+2x)=64

(x+1)^2=64/(2x+1)

решу его графически- точка пересечения графиков левой и правой части и есть решение

x≈2.3

Answer 2

To solve the equation log2 + log4 = 3, we can use logarithm properties to combine the logarithms into a single expression and then solve for x.

Recall the following logarithm properties:

1. loga + loga = log[a](b * c)
2. loga^m = m * loga

Using these properties, we can rewrite the equation as follows:

log2 + log4 = log2 + (log2^2) = log2 + (2 * log2).

Now the equation becomes:

log2 + 2 * log2 = 3.

Next, we can combine the logarithms:

log2 + 2 * log2 = log[2]((x+1) * (1+2x)^2).

So, the equation is now:

log[2]((x+1) * (1+2x)^2) = 3.

To get rid of the logarithm, we can rewrite the equation in exponential form:

2^3 = (x+1) * (1+2x)^2.

Simplify the equation:

8 = (x+1) * (1+2x)^2.

Now, let's solve for x:

Expand the right side of the equation:

8 = (x + 1) * (1 + 4x + 4x^2).

Now, rewrite the equation as a quadratic equation:

8 = 4x^3 + 5x^2 + 4x + 1.

Move all the terms to one side to set the equation to zero:

4x^3 + 5x^2 + 4x + 1 - 8 = 0.

Combine like terms:

4x^3 + 5x^2 + 4x - 7 = 0.

At this point, we have a cubic equation. Unfortunately, there is no straightforward algebraic way to solve cubic equations. To find the exact solutions for x, we would need to use numerical methods or a calculator that can handle polynomial equations.

Alternatively, you can try to find approximate solutions for x using numerical methods like the Newton-Raphson method or graphing the equation and finding the x-intercepts.