Solve Complex Logarithmic Equation: log₅9(log₃4x + log₃(4x+1)) = 2(log₅4a³ - log₅2a)

$\log_59(\log_34x+\log_3(4x+1))=2(\log_54a^3-\log_52a)$

Given a>0 , find X and express by a

The given problem requires solving the logarithmic equation $\log_5(9(\log_3(4x) + \log_3(4x + 1))) = 2(\log_5(4a^3) - \log_5(2a))$. We need to find $x$ in terms of $a$.

**Step 1:** Simplifying the left side using the product rule:

• $\log_3(4x) + \log_3(4x + 1) = \log_3((4x)(4x + 1)) = \log_3(16x^2 + 4x)$

**Step 2:** The equation becomes $\log_5(9 \log_3(16x^2 + 4x))$. To simplify, recognize $\log_5(9) + \log_5(\log_3(16x^2 + 4x))$.

**Step 3:** Now simplify the right-hand side:

• $2(\log_5(4a^3) - \log_5(2a)) = 2(\log_5(\frac{4a^3}{2a})) = 2(\log_5(2a^2)) = 2(\log_5(2) + \log_5(a^2))$
• $= 2 \log_5(2) + 2 \log_5(a^2) = 2 \log_5(2) + 4 \log_5(a) = 2 + 4 \log_5(a)$ (since $\log_5(2) = 1$)

**Step 4:** Equate both sides:

• $\log_5(9) + \log_5(\log_3(16x^2 + 4x)) = 2 + 4 \log_5(a)$

**Step 5:** Exponentiate and solve for $x$:

• Convert back from form: $9 \log_3(16x^2 + 4x) = 5^{2 + 4 \log_5(a)}$
• Further simplified using algebraic manipulation, and solve the quadratic in terms of $x$:
• $16x^2 + 4x = 5^{2 + 4 \log_5(a)}/9$
• Set: $x = -\frac{1}{8} + \frac{\sqrt{1 + 8a^2}}{8}$

Thus, the solution to the problem, and hence the expression for $x$ in terms of $a$, is:

$x = -\frac{1}{8} + \frac{\sqrt{1 + 8a^2}}{8}$.