How do you solve $ln (7 x + 6) = 1$ $ln(7x+6)=1$ ?

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How do you solve $ln (7 x + 6) = 1$ $ln(7x+6)=1$ ?

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Answer 1 · 2015-12-04T13:32:05

I found: $x = \frac{e - 6}{7}$ $x=(e-6)/7$

Explanation:

You can use the definition of log remembering that the base of $ln$ $ln$ is the irrational number $e$ $e$ :
$ln (7 x + 6) = 1$ $ln(7x+6)=1$
$7 x + 6 = e^{1}$ $7x+6=e^1$
rearranging:
$x = \frac{e - 6}{7}$ $x=(e-6)/7$

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How do you solve $ln (7 x + 6) = 1$ $ln(7x+6)=1$ ?

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How do you solve ln(7x+6)=1ln(7x+6)=1ln(7x+6)=1?

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How do you solve $ln (7 x + 6) = 1$ $ln(7x+6)=1$ ?