Logarithms: rewrite the equation for I, solving for t

[wind]

Hi, can some one check over my work? Thanks

\(\displaystyle \L \ I\, =\, 0.8(1\, -\, 10^{-0.0434t})\)

Instructions: Rewrite the equation for I, solving for t.

\(\displaystyle \L \ log I\, =\, log 0.8\, +\, log (1\, -\, 10^{-0.0434t})\)
.
\(\displaystyle \L \ log I\, =\, log 0.8\, -\, 0.0434t log (1\, -\, 10)\)
.
\(\displaystyle \L \ log I\, =\, log 0.8\, -\, 0.0434t log (9)\)
.
\(\displaystyle \L \ log I\, -\, log 0.8\, =\, -0.0434t\, log (9)\)
.
\(\displaystyle \L \frac{log I\, -\, log 0.8}{(log9)(-0.0434)} \, =\, t\)

 

[skeeter]

looks like dealing with logarithms is giving you fits ...

\(\displaystyle \log{(1 - 10^{-0.0434t})} = -0.0434\log{(1-10)}\) ???
I don't think so.

first off, let's rename those ugly constants to make it easier to keep track ...

let a = 0.8
let k = 0.0434

\(\displaystyle \L I = a(1 - 10^{-kt})\)

\(\displaystyle \L I = a - a \cdot 10^{-kt}\)

\(\displaystyle \L I - a = -a \cdot 10^{-kt}\)

\(\displaystyle \L a - I = a \cdot 10^{-kt}\)

\(\displaystyle \L \frac{a - I}{a} = 10^{-kt}\)

\(\displaystyle \L \log{\left(\frac{a-I}{a}\right)} = -kt\)

\(\displaystyle \L \frac{1}{k}\log{\left(\frac{a}{a-I}\right)} = t\)

now you can back-substitute those ugly constants if you wish.