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assignment 6.4: logarithmic equations and inequalities
score: 4/19 answered: 4/16
question 5
solve:
\\(\log_{3} (4x + 13) = \log_{3} (-7x + 13)\\)
\\(x = \\) (enter dne if no solution exists)

Explanation:

Step1: Set arguments equal

Since $\log_b(a) = \log_b(c)$ implies $a=c$ (for valid arguments), set $4x+13=-7x+13$.

Step2: Isolate x terms

Add $7x$ to both sides: $4x+7x+13=13$
$\implies 11x+13=13$

Step3: Solve for x

Subtract 13 from both sides: $11x=0$
$\implies x=\frac{0}{11}=0$

Step4: Verify validity

Check arguments:
$4(0)+13=13>0$, $-7(0)+13=13>0$. Both are valid for $\log_3$.

Answer:

$0$