Question 892191
There is no real result, but maybe there is a typo.
1408=246(1-e^(-9.61*10^-7t)) does not look good as {{{1408=246(1-e^(-9.61*10^-7t))}}} ,
but I could write it as {{{1408=246(1-e^(-0.000000961*t))}}}
To solve, the equation could be transformed to
{{{1408=246-246*e^(-0.000000961*t))}}}-->{{{1408-246=-246*e^(-0.000000961*t))}}}-->{{{1162=-246*e^(-0.000000961*t))}}}
That cannot be true for any value of the variable because exponential functions are always positive: {{{e^anything>0}}} .
That would make the right side of the equation negative, and the left side is positive.
 
Curves of the type {{{f(t)=A(1-e^(-k*t)))}}}
Have {{{f(0)=0}}} and asymptotically approach {{{A}}} as {{{t}}} increases.
If the equation was {{{1408=2460*(1-e^(-0.000000961*t)))}}} ,
then the graph of {{{f(t)=2460(1-e^(-0.000000961*t)))}}} would look like this: {{{graph(300,300,-0.2,0.8,-300,2700,2460*(1-e^(-5x)),2460)}}}
and we could solve it like this:
{{{1408=2460-2460*e^(-0.000000961*t)}}}
 
{{{1408=2460-2460*e^(-0.000000961*t)}}}
 
{{{1408-2460=-2460*e^(-0.000000961*t)}}}
 
{{{-1052=-2460*e^(-0.000000961*t))}}}
 
{{{1052=2460*e^(-0.000000961*t))}}}
 
{{{1052/2460=e^(-0.000000961*t))}}}
 
{{{0.42764=e^(-0.000000961*t))}}}
 
{{{ln(0.42764)=-0.000000961*t}}}
 
{{{-.84947=-0.000000961*t}}}
 
{{{(-.84947)/(-0.000000961)=t}}} --> {{{t=8.84*10^5}}}