To solve the equation:

$\log_8(x + 7) = \log_3(4x)$

we first apply the property of logarithms that states if:

$\log_a(M) = \log_a(N)$

then:

$M = N$

Therefore, we set the expressions inside the logarithms equal to each other:

$x + 7 = 4x$

Now, solve for $x$:

1. Subtract $x$ from both sides: $7 = 3x$

2. Divide by 3: $x = \frac{7}{3}$

It seems none of the given options match $x = \frac{7}{3}$, but let’s verify whether this solution works in the original equation.

The solution $x = \frac{7}{3}$ makes neither side of the equation meaningful in this context because neither the base 8 nor base 3 logarithms can take negative or zero values for their arguments.

Since no valid value satisfies the equation under real numbers and no correct answer fits, the equation may have no real solutions, though rechecking the options can clarify.

Would you like more details on the logarithmic approach or why no solutions match?

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Related Questions:

1. What are the key properties of logarithms used to simplify such equations?
2. How can logarithms with different bases (such as 8 and 3) be handled?
3. What does it mean for a logarithmic argument to be valid?
4. Can equations with logarithms of different bases have real solutions?
5. How can you graph logarithmic functions to visually verify solutions?

Tip:

Always check that the arguments of logarithms are positive before solving such equations!