log_(5)(10x)-1=log_(5)(3x-1) 4. $$log_5 (10x) - 1 = log_5 (3x - 1)$$
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For $log_5 (10x)$, we must have $10x > 0$, which implies $x > 0$. For $log_5 (3x - 1)$, we must have $3x - 1 > 0$, which implies $3x > 1$, so $x > \frac{1}{3}$. Combining these conditions, we must have $x > \frac{1}{3}$. Show more…
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