Solve for x log base 7 of x^2-5- log base 7 of x+1=0
The question asks you to solve an equation involving logarithms with base 7. Specifically, you need to find the value(s) of the variable x that satisfy the equation given by the difference of two logarithmic expressions: the logarithm base 7 of x squared minus 5, minus the logarithm base 7 of x plus 1, equating to 0. To solve for x, you'll need to manipulate the equation using logarithmic properties and potentially other algebraic methods.
$\left(log\right)_{7} \left(\right. x^{2} - 5 \left.\right) - \left(log\right)_{7} \left(\right. x + 1 \left.\right) = 0$
Answer
Solution:
Step 1:
Apply the quotient rule of logarithms, which states that $\log_b(x) - \log_b(y) = \log_b\left(\frac{x}{y}\right)$. Thus, we get $\log_7\left(\frac{x^2 - 5}{x + 1}\right) = 0$.
Step 2:
Convert the logarithmic equation into its equivalent exponential form. Recall that if $\log_b(x) = y$, then $b^y = x$. Therefore, $7^0 = \frac{x^2 - 5}{x + 1}$.
Step 3:
Eliminate the fraction by cross-multiplying to get $x^2 - 5 = 7^0(x + 1)$.
Step 4:
Simplify the right side of the equation.
Step 4.1:
Recognize that any number raised to the power of $0$ is $1$, so we have $x^2 - 5 = 1(x + 1)$.
Step 4.2:
Multiply out the right side to get $x^2 - 5 = x + 1$.
Step 5:
Subtract $x$ from both sides to obtain $x^2 - x - 5 = 1$.
Step 6:
Rearrange the equation by moving constants to the other side.
Step 6.1:
Add $5$ to both sides to get $x^2 - x = 6$.
Step 6.2:
Combine like terms to maintain the equation $x^2 - x = 6$.
Step 7:
Subtract $6$ from both sides to form a quadratic equation $x^2 - x - 6 = 0$.
Step 8:
Factor the quadratic equation.
Step 8.1:
Find two numbers that multiply to $-6$ and add to $-1$. The numbers are $-3$ and $2$.
Step 8.2:
Express the factored form as $(x - 3)(x + 2) = 0$.
Step 9:
Set each factor equal to zero, since if either factor is zero, the product is zero.
Step 10:
Solve the first factor for $x$.
Step 10.1:
Set $x - 3$ equal to $0$ to get $x - 3 = 0$.
Step 10.2:
Add $3$ to both sides to find $x = 3$.
Step 11:
Solve the second factor for $x$.
Step 11.1:
Set $x + 2$ equal to $0$ to get $x + 2 = 0$.
Step 11.2:
Subtract $2$ from both sides to find $x = -2$.
Step 12:
Combine the solutions to get $x = 3$ or $x = -2$.
Step 13:
Exclude any solutions that do not satisfy the original logarithmic equation. The valid solution is $x = 3$.
Knowledge Notes:
Quotient Rule of Logarithms: The quotient rule states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator: $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$.
Exponential Form of a Logarithm: The exponential form is derived from the definition of a logarithm: if $\log_b(x) = y$, then $b^y = x$.
Cross Multiplication: This is a method used to solve equations involving fractions. By multiplying both sides of the equation by the denominators, we can eliminate the fractions.
Factoring Quadratic Equations: To factor a quadratic equation of the form $ax^2 + bx + c = 0$, one must find two numbers that multiply to $ac$ and add up to $b$. These numbers are then used to split the middle term and factor by grouping.
Zero Product Property: If a product of factors equals zero, then at least one of the factors must be zero. This property is used to solve quadratic equations after factoring.
Excluding Extraneous Solutions: In the context of logarithmic equations, it's important to exclude any solutions that would make the argument of the logarithm negative or zero, as the logarithm is not defined for these values.
