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lny=9t-8; solve for y?
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\[y=e^{9t-8}\]
just rewrite in equivalent exponential form
What's that?
y = e^(9t-8)
\[\ln(\heartsuit)=\spadesuit+\iff e^{\spadesuit}=\heartsuit\]
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you should be able to switch from one form to the other easily. it is the same saying \[\log_b(x)=y\iff b^y=x\]
y=e^(9t-8)
in this case \[\ln(x)=\log_e(x)\]
I still don't really understand, so would the form be 4.09? Like if you solved it? Or just the y= equation you gave me?
Solving for 'y' just means that you have that variable isolated in one side of the equation (by convention, the left side). Since no values for 't' are specified, the solution posted by satellite73 is the answer IMHO.
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