How do you solve $ln (5 x - 4) = ln 2 + ln (x + 1)$ $Ln(5x-4)=ln2+ln(x+1)$ ?

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How do you solve $ln (5 x - 4) = ln 2 + ln (x + 1)$ $Ln(5x-4)=ln2+ln(x+1)$ ?

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Answer 1 · 2016-05-13T16:19:30

$x = 2$ $x=2$

Explanation:

Taking $ln (5 x - 4) = ln (2) + ln (x + 1)$ $ln(5x-4) = ln(2)+ln(x+1)$ or equivalently $ln (5 x - 4) - ln (2 \cdot (x + 1)) = 0$ $ln(5x-4)-ln(2*(x+1))=0$ or $ln [\frac{5 x - 4}{2 (x + 1)}] = ln (1)$ $ln[(5x-4)/(2(x+1))]= ln(1)$ then results in
$\frac{5 x - 4}{2 (x + 1)} = 1$ $(5x-4)/(2(x+1))=1$ or $5 x - 4 = 2 (x + 1)$ $5x-4=2(x+1)$ and finally $x = 2$ $x = 2$

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How do you solve $ln (5 x - 4) = ln 2 + ln (x + 1)$ $Ln(5x-4)=ln2+ln(x+1)$ ?

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How do you solve Ln(5x-4)=ln2+ln(x+1)ln(5x−4)=ln2+ln(x+1)Ln(5x-4)=ln2+ln(x+1)?

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How do you solve $ln (5 x - 4) = ln 2 + ln (x + 1)$ $Ln(5x-4)=ln2+ln(x+1)$ ?