<p>How can I solve for e in this problem without graphing stuff on my ti-83? I've tried everything and can't seem to get the ln y or the x on its own...</p>
<p>e^x + e^-x = 3</p>
<p>ln(y-1) - ln(2) = x + ln x,
and trying to solve for y.</p>
<p>Are these typical SAT II math problems?</p>
<p>well you can go
ln(y-1)-ln(2) =
ln((y-1)/2) = x + ln(x)
(y-1)/2 = e^((x) +ln(x))
= e^x * e^lnx
= e^x *x
y = 2xe^x + 1</p>
<p>Try learning rules of exponents for the 1st one.</p>
<p>e^x + e^-x = 3</p>
<p>Multiply e^x to both sides.
Then sub y = e^x.
Solve the quadratic eqn.</p>
<p>buy a TI-89 Titanium.</p>
<p>:-D</p>
<p>Yep. I don't know anything about natural log...I just learned how to imput it on my calculator. Got through the test just fine.</p>