What is the solution to the logarithmic equation log₇(x) + log₇(x + 1) = log₇(6)?

What is the solution to the logarithmic equation log₇(x) + log₇(x + 1) = log₇(6)?

• A. 1
• B. 2
• C. 3
• D. 4

Answer: (B)

To solve the equation \( log₇(x) + log₇(x + 1) = log₇(6) \), we can first use the properties of logarithms, specifically the property that states \( log_a(b) + log_a(c) = log_a(b \cdot c) \). By applying this property, we can combine the left-hand side: \[ log₇(x) + log₇(x + 1) = log₇(x \cdot (x + 1)) \] Thus, the equation simplifies to: \[ log₇(x \cdot (x + 1)) = log₇(6) \] Since the logarithmic function is one-to-one, we can remove the logarithms from both sides: \[ x \cdot (x + 1) = 6 \] Expanding the left side gives us: \[ x^2 + x = 6 \] Rearranging this leads to a standard quadratic equation: \[ x^2 + x - 6 = 0 \] We can factor this quadratic: \[ (x - 2)(x + 3) = 0 \]

To solve the equation ( log₇(x) + log₇(x + 1) = log₇(6) ), we can first use the properties of logarithms, specifically the property that states ( log_a(b) + log_a(c) = log_a(b \cdot c) ). By applying this property, we can combine the left-hand side:

[

log₇(x) + log₇(x + 1) = log₇(x \cdot (x + 1))

]

Thus, the equation simplifies to:

[

log₇(x \cdot (x + 1)) = log₇(6)

]

Since the logarithmic function is one-to-one, we can remove the logarithms from both sides:

[

x \cdot (x + 1) = 6

]

Expanding the left side gives us:

[

x^2 + x = 6

]

Rearranging this leads to a standard quadratic equation:

[

x^2 + x - 6 = 0

]

We can factor this quadratic:

[

(x - 2)(x + 3) = 0

]