How do you solve $ln (4 x + 1) - ln 3 = 5$ $ln(4x+1)-ln3=5$ ?

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Answer 1 · 2016-11-11T18:13:39

Please see the explanation for steps leading to: $x = \frac{3 e^{5} - 1}{4}$ $x = (3e^5 - 1)/4$

Use the property of logarithms ${log}_{b} (x) - {log}_{b} (y) = {log}_{b} (\frac{x}{y})$ $log_b(x) - log_b(y) = log_b(x/y)$ :

$ln (\frac{4 x + 1}{3}) = 5$ $ln((4x + 1)/3) = 5$

Write both sides as exponents of e:

$e^{ln (\frac{4 x + 1}{3})} = e^{5}$ $e^{ln((4x + 1)/3)} = e^5$

Use the property $e^{ln (a)} = a$ $e^ln(a) = a$ :

$\frac{4 x + 1}{3} = e^{5}$ $(4x + 1)/3 = e^5$

Multiply both side by 3:

$4 x + 1 = 3 e^{5}$ $4x + 1 = 3e^5$

Subtract 1 from both sides:

$4 x = 3 e^{5} - 1$ $4x = 3e^5 - 1$

Divide both sides by 4:

$x = \frac{3 e^{5} - 1}{4}$ $x = (3e^5 - 1)/4$

How do you solve ln(4x+1)-ln3=5ln(4x+1)−ln3=5ln(4x+1)-ln3=5?