Problem 135 $$ \log (x-3)+\log (x+6)=1 $... [FREE SOLUTION]

Chapter 7: Problem 135

$$ \log (x-3)+\log (x+6)=1 $$

Short Answer

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The short answer is: To solve the given logarithmic equation $\log (x-3) + \log (x+6) = 1$, we first combined the logarithms into a single logarithm $\log [(x-3)(x+6)] = 1$, then removed the logarithm using exponentiation to get $(x-3)(x+6) = 10$. Next, we expanded and simplified the equation to $x^2 + 3x - 28 = 0$ which factors as $(x + 7)(x - 4) = 0$. Solving for x, we found two possible solutions: $x = -7$ and $x = 4$. However, since the x = -7 solution results in a negative argument for the first logarithm, we discarded it. The only valid solution is $x = 4$.

Step by step solution

Combine the logarithms

Using the logarithm property $\log ab = \log a + \log b$, we can combine the two logarithms: $\log (x-3) + \log (x+6) = \log [(x-3)(x+6)]$ The equation now becomes: $\log [(x-3)(x+6)] = 1$

Remove the logarithm using exponentiation

To remove the logarithm, we can use the exponentiation property. Since the equation is in the form $y = \log_b x$, we can rewrite it as $x = b^y$. Here, the base, b, is 10 since it is not specified: $(x-3)(x+6) = 10^1$ Now, the equation is: $(x-3)(x+6) = 10$

Solve for x

First, expand the left side of the equation: $x^2 + 6x - 3x -18 = 10$ Combine the like terms: $x^2 + 3x - 18 = 10$ Subtract 10 from both sides to set the equation to zero: $x^2 + 3x - 28 = 0$ Now, we can factor the quadratic equation: $(x + 7)(x - 4) = 0$ Finally, we can solve for x: $x + 7 = 0 \Rightarrow x = -7$ Or $x - 4 = 0 \Rightarrow x = 4$ However, we need to check if these values of x satisfy the original equation (i.e., they do not make the argument of any logarithm negative). The first solution, x = -7, would indeed result in a negative argument for the first logarithm, which is not possible. Therefore, we discard the x = -7 solution. The only valid solution is: $x = 4$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponentiation

In mathematics, exponentiation is the operation of raising one number, known as the base, to the power of another number, which is the exponent. It is a fundamental operation that is the opposite of taking a logarithm.
For example, if we have a base number of 10 and an exponent of 1, we write it as $10^1$, which equals 10. This operation was used in the step where we had $\log[(x-3)(x+6)] = 1$. Here, using an implied base of 10 (since no base was mentioned, it defaults to 10), we re-write the logarithmic equation by converting it into its exponential form: $(x-3)(x+6) = 10^1$, which simplifies to $(x-3)(x+6) = 10$.

Remember, exponentiation helps in converting equations to a form that can make solving easier.
It is especially useful in removing logarithms to simplify equations temporarily.

Quadratic Equations

Quadratic equations are polynomials of the second degree, typically in the form $ax^2 + bx + c = 0$. In our logarithmic equation solution, after using exponentiation, we landed with the equation $(x-3)(x+6) = 10$, which simplifies to $x^2 + 3x - 28 = 0$ upon expansion and combination of terms.
This is a standard quadratic equation in which:

$a = 1$
$b = 3$
$c = -28$

Once we set the quadratic equation to zero, $(x + 7)(x - 4) = 0$, we can use factoring to find the roots of the polynomial.
Quadratic equations can often have two solutions because they involve squaring (hence the term "quadratic," which refers to "square"). For our equation:

The roots are $x = -7$ and $x = 4$.
However, only $x = 4$ is valid within the context of our problem as $x = -7$ would make one of the logarithmic arguments negative, which is not allowed.

Properties of Logarithms

Logarithms are the inverse operation of exponentiation and have unique properties that make them useful in solving equations. In this particular problem, we used the property $\log(ab) = \log a + \log b$ to combine two separate logarithms into a single logarithmic expression:
$\log(x-3) + \log(x+6) = \log[(x-3)(x+6)]$.
This simplification step is crucial as it allows us to transform the equation into a form that can be tackled using exponentiation.

One key property of logarithms is that they allow us to solve exponential equations by converting them into linear or polynomial equations.
Another useful property is the change of base formula, although not used here, it helps in computing logarithms with different bases.

By understanding and using such properties, complex equations can be simplified and solved more easily, as we demonstrated in this problem with the property $\log(ab) = \log a + \log b$.

91影视

$$ \log (x-3)+\log (x+6)=1 $$

Short Answer

Step by step solution

Combine the logarithms

Remove the logarithm using exponentiation

Solve for x

Key Concepts

Exponentiation

Quadratic Equations

Properties of Logarithms

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