Let’s solve the equation step by step. The given equation is:

$\log_2(x+7) = -\log_2(x+5) + 3$

Step 1: Combine logarithmic terms

We use the property of logarithms: $-\log_b(a) = \log_b(a^{-1})$, so the equation becomes:

$\log_2(x+7) = \log_2\left(\frac{1}{x+5}\right) + 3$

Step 2: Isolate the logarithmic terms

Subtract $\log_2\left(\frac{1}{x+5}\right)$ from both sides:

$\log_2(x+7) - \log_2\left(\frac{1}{x+5}\right) = 3$

Using the property $\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$:

$\log_2\left(\frac{x+7}{\frac{1}{x+5}}\right) = 3$

Simplify the fraction:

$\log_2\left((x+7)(x+5)\right) = 3$

Step 3: Rewrite in exponential form

Using the definition of a logarithm, $y = \log_b(x) \implies b^y = x$:

$(x+7)(x+5) = 2^3$

Simplify $2^3 = 8$:

$(x+7)(x+5) = 8$

Step 4: Expand and simplify

Expand the left-hand side:

$x^2 + 5x + 7x + 35 = 8$

Combine like terms:

$x^2 + 12x + 35 = 8$

Simplify further:

$x^2 + 12x + 27 = 0$

Step 5: Solve the quadratic equation

We solve $x^2 + 12x + 27 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 12$, $c = 27$. Substitute into the formula:

$x = \frac{-12 \pm \sqrt{12^2 - 4(1)(27)}}{2(1)}$

$x = \frac{-12 \pm \sqrt{144 - 108}}{2}$

$x = \frac{-12 \pm \sqrt{36}}{2}$

$x = \frac{-12 \pm 6}{2}$

Solve for both cases:

1. $x = \frac{-12 + 6}{2} = \frac{-6}{2} = -3$
2. $x = \frac{-12 - 6}{2} = \frac{-18}{2} = -9$

Step 6: Check for extraneous solutions

For a logarithmic equation, the argument of the logarithm must be positive:

1. For $x = -3$: $x+7 = 4 > 0$, $x+5 = 2 > 0$ → Valid.
2. For $x = -9$: $x+7 = -2$ (invalid, as the argument is negative).

Thus, the only solution is:

$x = -3$

Would you like more details or have any questions?

Related Questions:

1. What are the properties of logarithms used in this problem?
2. How do you check for extraneous solutions in logarithmic equations?
3. What is the quadratic formula and when do we use it?
4. Why must the argument of a logarithm always be positive?
5. How can we apply similar steps to solve exponential equations?

Tip:

Always verify potential solutions in logarithmic equations to ensure the arguments remain valid!