How do you solve $4 e^{9 x - 1} = 64$ $4e^(9x-1)=64$ ?

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Answer 1 · 2016-03-08T12:21:14

$x = \frac{ln (16) + 1}{9}$ $x=(ln(16)+1)/9$

Divide both sides of the equation first. This will deal with the $4$ $4$ attached to the $e^{9 x - 1}$ $e^(9x-1)$ .

$e^{9 x - 1} = 16$ $e^(9x-1)=16$

Now, take the natural logarithm of both sides of the equation.

$ln (e^{9 x - 1}) = ln (16)$ $ln(e^(9x-1))=ln(16)$

Note that the $ln$ $ln$ and $e^{x}$ $e^x$ functions are inverses, so they undo one another.

$9 x - 1 = ln (16)$ $9x-1=ln(16)$

Solving through basic algebra,

$x = \frac{ln (16) + 1}{9}$ $x=(ln(16)+1)/9$

How do you solve 4e9x−1=644e^(9x-1)=64?