How do you solve $ln((e^(4x+3))/e)=1$ ?

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Answer 1 · 2015-12-15T09:52:09

$x = -1/4$

Use the following logarithmic law first:

$ln (a/b) = ln(a) - ln(b)$

In your case, this leads to:

$ln(e^(4x+3)/e) = 1$

$<=> ln(e^(4x+3)) - ln(e) = 1$

As next, you need to use the property that $ln x$ and $e^x$ are inverse functions which means that $ln(e^x) = x$ and $e^(ln x) = x$ always hold.

Thus, $ln$ and $e$ eliminate each other in your equation, and you will get:

$<=> (4x + 3) - 1 = 1$

The solution of this equation is

$x = -1/4$

As $e^x$ is always positive for any value of $x in RR$ , and thus the logarithmic expression is defined for any $x in RR$ , this is your solution.

How do you solve ln((e^(4x+3))/e)=1ln((e^(4x+3))/e)=1?