Question

Geogebra said this function is: 8.63 e^(-0.53x)
F(x)=8.63e^(-0.53x)
Y=8.63e^(-0.53x)
In y/8.63= In e^(-0.53x)
In y – In 8.63= In e^(-0.53x)
In y= -0.53x + 2.16
Then it said change it into half-life function but
I don't get how I am supposed to write a half life function out of that
Half life formula: A=[1/2]^x

Answer 1

logs are confusing

but you could say this:

e to some power = 1/2
by definition, this is
\[ e^{ln(1/2)} = \frac{1}{2} \]
if you raise both sides to the 1/ln(1/2)
\[ e^{\frac{ln(1/2)}{\ln(1/2)}} = \left(\frac{1}{2}\right)^{\frac{1}{\ln(1/2)}} \]

this last step used the idea that
\[ (e^x)^\frac{1}{x} = e^{\frac{x}{x}} = e^1= e\]

so you now know
\[ \left(\frac{1}{2}\right)^{\frac{1}{ln(1/2)}} = e\]

plot that mess into your equation instead of e:
\[ F(x)=8.63e^{-0.53x} \\
F(x)=\left(\left(\frac{1}{2}\right)^{\frac{1}{\ln(1/2)}}\right)^{-0.53x} \\
F(x)=\left(\frac{1}{2}\right)^{- \frac{0.53}{\ln(1/2)}x}
\]

Answer 2

**8.63 should still be in the last 2 equations

Answer 3

thankss:) @phi