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

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Answer 1 · 2016-05-18T18:13:16

We have that

$4 \cdot e^{x - 1} = 64$ $4*e^(x-1)=64$

$\frac{4 \cdot e^{x - 1}}{4} = \frac{64}{4}$ $(4*e^(x-1))/4=64/4$ (Divide by 4 both sides)

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

$(x - 1) \cdot ln e = ln 16$ $(x-1)*lne=ln16$ (Take logarithms in both sides)

$(x - 1) = 4 \cdot ln 2$ $(x-1)=4*ln2$ (Remember that $16 = 2^{4}$ $16=2^4$ )

$x = 1 + 4 \cdot ln 2$ $x=1+4*ln2$

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