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Mathematics 7 Online
OpenStudy (anonymous):

What is the solution of the equation exe2x = 4?

OpenStudy (anonymous):

@austinL

OpenStudy (austinl):

Is it, \(e^{2x}=4\)?

hartnn (hartnn):

or \(\large e^x e^{2x}=4 \\?\)

OpenStudy (anonymous):

@hartnn yeh that

OpenStudy (austinl):

I would rewrite it as, \(e^{x}\times e^{2x}=4\) Then take the natural log of each side.

OpenStudy (anonymous):

so x * 2x = 4

OpenStudy (anonymous):

wouldnt that make x squared ?

OpenStudy (austinl):

Yep, we would have, \(2x^2=\ln(4)\)

OpenStudy (anonymous):

would that be my final answer ? or would it be \[x = \sqrt{ \frac{ \ln4 }{ 2}}\]

OpenStudy (austinl):

That is close, you need to simplify that. \(\displaystyle x=\frac{\sqrt{\ln(4)}}{\sqrt{2}}\) Do you know how to do that?

OpenStudy (anonymous):

no /:

OpenStudy (anonymous):

can i just plug it into my calculator ?

OpenStudy (austinl):

You need to multiply both top and bottom by \(\sqrt{2}\)

OpenStudy (anonymous):

i got x = .832 is that right ?

OpenStudy (austinl):

Remember what I said, does it tell you to give a decimal approximation? Do you have choices?

OpenStudy (anonymous):

no i have to write out my answer. its an online class

hartnn (hartnn):

wait what ?

OpenStudy (anonymous):

@hartnn wym ?

hartnn (hartnn):

\(\huge x^m \times x^n = x^{m+n}\)

OpenStudy (austinl):

It is \(x\times2x=2x^2\) @hartnn

hartnn (hartnn):

\(\huge e^x \times e^{2x} = e^{x+2x}\)

hartnn (hartnn):

x+2x = ln 4 x = (ln 4) /3

OpenStudy (anonymous):

im soooo lost

OpenStudy (austinl):

Oh my goodness..... I can't believe I made that error...

OpenStudy (austinl):

hartnn is correct.

OpenStudy (anonymous):

where do i have to go back and check ?

hartnn (hartnn):

which step did u not get ?

OpenStudy (anonymous):

you confused me ,

OpenStudy (anonymous):

okay so ln on both sides gives me x * 2x = 4ln right ?

OpenStudy (anonymous):

ln 4 i mean

hartnn (hartnn):

how x*2x ?

OpenStudy (austinl):

I made an error, it is, x+2x=ln(4) 3x=ln(4)

OpenStudy (austinl):

x=ln(4)/3

OpenStudy (anonymous):

why plus ?

hartnn (hartnn):

\(\huge x^m \times x^n = x^{m+n}\)

OpenStudy (austinl):

Because I made an error. Go back and look at hartnn's posts.

OpenStudy (anonymous):

oh i see

OpenStudy (anonymous):

so 3x = 4ln

hartnn (hartnn):

so e^x e^2x = e^ (x+2x) only then you take natural log on both sides

hartnn (hartnn):

3x = ln 4

OpenStudy (anonymous):

i take natural log to bring the x's down right ?

hartnn (hartnn):

yes

hartnn (hartnn):

and the fact that ln e =1

OpenStudy (anonymous):

so my answer is ln 4 / 3 and i have the choice to make it a decimal im guessing ?

OpenStudy (austinl):

We have, \(\large{e^{x}\times e^{2x}=4\Rightarrow e^{x+2x}=4}\) Then you take the natural log, \(\large{x+2x=\ln(4)\Rightarrow3x=\ln(4)}\) \(\large{\displaystyle x=\frac{\ln(4)}{3}}\)

OpenStudy (austinl):

I would leave it as is, that is the exact answer.

OpenStudy (austinl):

Good catch @hartnn I made a silly error and you saw it, I would have guided them along the wrong path.... Silly me.

hartnn (hartnn):

no problem, thats what we call team work :)

OpenStudy (anonymous):

wow thanks so muuch !

OpenStudy (anonymous):

highly appreciate it

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