Question

Log2(x+1)+log2(x+3)=3 log3x2-log3x3=-2

Answer 1

$\log_2(x+1)+\log_2(x+3)=3,\quad x \in (-1, + \infty)\\\log_2((x+1)\times(x+3))=3\\\log_2(x^2+3x+x+3)=3\\x^2+3x+x+3=2^3\\x^2+4x+3-8=0\\x^2+5x-x-5=0\\x(x+5)-(x-5) = 0\\(x+5)(x-1)=0\\x+5=0\\x-1=0\\x_1 = -5 \\x_2=1\ x \in (-1,+\infty)\\\\\fbox{x=1}$

$\log_3(x^2)-\log_3(x^3)=-2,\quad x\in(0,+\infty)\\2\log_3(x)+3\log_3(x)=-2\\5\log_3(x)=-2\\\log_3(x) = -\dfrac{2}{5}\\x = 3^ {-\dfrac{2}{5}}\\\\ x = \dfrac{\sqrt[5]{3^3}}{3} }=\dfrac{\sqrt[5]{27}}{3} }$

Answer 2

To solve the given equations, let's start with the first equation:

log2(x + 1) + log2(x + 3) = 3

We can use the logarithmic properties to simplify this equation. According to the logarithmic property, the sum of logarithms is equal to the logarithm of the product:

log2[(x + 1)(x + 3)] = 3

Now we can rewrite the equation in exponential form:

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

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

Expanding the right side of the equation:

8 = x^2 + 4x + 3

Rearranging the terms:

x^2 + 4x + 3 - 8 = 0

x^2 + 4x - 5 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula. Factoring the equation, we find:

(x + 5)(x - 1) = 0

This gives us two possible solutions:

x + 5 = 0 => x = -5 x - 1 = 0 => x = 1

So the solutions to the first equation are x = -5 and x = 1.

Moving on to the second equation:

log3(x^2) - log3(x^3) = -2

Using the logarithmic property, we can simplify the equation:

log3(x^2 / x^3) = -2

log3(1/x) = -2

Now, let's convert this equation to exponential form:

3^(-2) = 1/x

1/9 = 1/x

Cross-multiplying:

9x = 1

x = 1/9

Therefore, the solution to the second equation is x = 1/9.

To summarize, the solutions to the given equations are:

• x = -5
• x = 1
• x = 1/9