Question

Can anyone help me with this trig question (reviewing for a finals test)

e^x + 16e^-x
---------------- = 4
2

Answer 1

So umm... multiply by 2,\[\large\rm e^x+16e^{-x}=8\]Multiply by e^x to deal with this negative power,\[\large\rm (e^x)^2+16=8e^x\]Any confusion there?

Answer 2

So so you need to multply by e^x to take out the negative?

Ok, I understand that part.

Answer 3

Yes, because \(\large\rm e^x\cdot e^{-x}=e^{x-x}=e^0=1\)

We'll subtract 8e^x to the other side,\[\large\rm (e^x)^2-8e^x+16=0\]So what we've done so far is, turned out expression into a quadratic.

Answer 4

And luckily, this one will factor.

Answer 5

\[\large\rm (\color{orangered}{e^x})^2-8\color{orangered}{e^x}+16=0\]If the exponential is confusing you,
you can temporarily replace it with something else, let's say u,\[\large\rm \color{orangered}{u}^2-8\color{orangered}{u}+16=0\]From this point, remember how to factor?

Answer 6

(u-4)^2

u = 4

ln4
---- correct?
ln 8

Answer 7

\[\large\rm (\color{orangered}{u}-4)^2=0\]Mmm k good.
So then,\[\large\rm \color{orangered}{u}=4\qquad\qquad\to\qquad\qquad \color{orangered}{e^x}=4\]And you take natural log to get the x out of the exponent position,\[\large\rm \ln(e^x)=\ln4\qquad\qquad\to\qquad\qquad x=\ln4\]I'm not sure where that ln(8) is coming from though.. hmm

Answer 8

no wait you're right,

thanks!

Answer 9

np

Answer 10

\[\large\rm 3^{2x} + 3^x - 42 = 0\]Written as,\[\large\rm (3^x)^2+3^x-42=0\]which you chose to write as,\[\large\rm u^2+u-42=0\]factoring into,\[\large\rm (u-6)(u+7)=0\]So we have,\[\large\rm 3^x=6\qquad\qquad\qquad 3^x=-7\]Out exponential can never equal a negative,
so we can ignore the -7.

If you chose to take `natural log`,\[\large\rm \ln(3^x)=\ln6\]Applying one of our log rules,\[\large\rm x\ln(3)=\ln6\]Divide by ln3,\[\large\rm x=\frac{\ln6}{\ln3}\]

Answer 11

In the previous problem what had happened was,
when we took our natural log,\[\large\rm \ln(e^x)=\ln4\]Again applying our log rule,\[\large\rm x\ln(e)=\ln4\]But in this case, ln(e)=1,
so we can ignore it,\[\large\rm x\cdot1=\ln4\]

Answer 12

oh the difference was that that ln4 never had a number other one, so there was no need for a deonominator

correct?

Answer 13

In the previous problem we could, from this step,\[\large\rm x\ln(e)=\ln4\]Divided by ln(e),\[\large\rm x=\frac{\ln4}{\ln e}\]And now it has the exact same format as that other problem, right?
But again, ln(e)=1.\[\large\rm x=\frac{\ln4}{1}\qquad\qquad\to\qquad\qquad x=\ln4\]

Answer 14

The difference was that the "other number" was "special",
it matched the base of the log.
(Base of natural log is e)

Answer 15

right, thanks!