pre-cal & trig: e^x + e^-x = 3, 2^x + 2^-x = 5, etc

[myimmortalfire]

Can someone please help me to solve three trig problems.

Problems:

1) e^x + e^-x = 3

2) 2^x + 2^-x = 5

3) ln(y - 1) - ln(2) = x + lnx

Thanks,
NSN

 

[stapel]

Can someone please help me to solve three trig problems.

Um... These aren't actually trigonometric equations; they're algebraic. :shock:

1) e^x + e^-x = 3

2) 2^x + 2^-x = 5

3) ln(y - 1) - ln(2) = x + lnx

Have you studied logarithms or exponentials at all? If not, we can provide links to lessons, so that you can learn the background material necessary to understand any hints we would provide.

(Note: Legitimate tutors don't generally "do" students' work for them, nor give out the answers. The volunteers here generally provide hints, links, examples, and instructions. But to understand what they provide, you need the foundational underpinnings.)

If you have studied these topics, then:

1) Multiply through by e^x, and then solve the resulting quadratic in e^x.

2) This works just like (1), except that you multiply through by 2^x.

3) This has two variables, so I'm guessing the instructions weren't to "solve". But what were the instructions?

Thank you!

Eliz.

P.S. Welcome to FreeMathHelp! :wink:

 

[myimmortalfire]

PROBLEM

Here is my work...

2) 2^x +2^-x = 5

when solving for x, i made the 2^-x a fraction (2^(1/x))
next i found the common denominator (aka 2^x)
then i multiplied the common denominator with the entire equation and ended up with an equation that looks like this:
2^2x - 5(2)^x + 1 = 0
now i used the quadratic formula to solve for x
my answer was 5 + or - the square root of 21 and the entire thing was divided by 2.

is this correct?

3) ln(y - 1) - ln(2) = x + ln(x)

ln((y-l)/2) = x (1 + ln1)

ln((y -1)/2) = x (l +0)

therefore x is ln((y -1)/2)

is this correct?

if you don't mind can you please send me a reply by the end of today, since my test is tomorrow.

sorry for the short notice.

 

[pka]

Re: PROBLEM

Here is my work...
2) 2^x +2^-x = 5 when solving for x, i made the 2^-x a fraction (2^(1/x))

That is incorrect!
\(\displaystyle \L 2^{ - x} = \frac{1}{{2^x }}\)

 

[myimmortalfire]

thanks

oops!!! typed it wrong the second one sorry were you able to look at the 3rd one

Thanks for all your help.
NSN

 

[stapel]

2) 2^x +2^-x = 5

...my answer was 5 + or - the square root of 21 and the entire thing was divided by 2.

To check the answer to any "solving" problem, plug the solution back in and see if it works:


. . .\(\displaystyle \L x\, =\, \frac{5\, +\, sqrt{21}}{2}\, \approx \, 4.8:\)

. . . . .\(\displaystyle \L 2^{\frac{5\pm\sqrt{21}}{2}}\, +\, 2^{-\frac{5\pm\sqrt{21}}{2}}\, \approx \,2^{4.8} + 2^{-4.8}\, \approx \, 27.9\)


This clearly doesn't work (not even within calculator round-off error), so your solution is probably not valid.

Note: I have a feeling that what you actually got was:


. . . . .\(\displaystyle \L 2^x\, =\, \frac{5\, \pm \, sqrt{21}}{2}\)


You now need to solve that for x.

3) ln(y - 1) - ln(2) = x + ln(x)

...therefore x is ln((y -1)/2) is this correct?

Since we still don't know the instructions, there is no way to tell. Sorry.

please send me a reply by the end of today

As was noted in the "Read Before Posting" thread you read before posting, "There is no paid staff waiting on-hand to give instant replies. Many of the volunteer tutors have 'real' jobs, and they all have to sleep from time to time. The people 'viewing' your posts may be fellow students. Please don't be offended if there are 'views' but no replies. It may take hours, even days, for a tutor, qualified in your topic's area, to respond." :wink:

Thank you for your consideration!

Eliz.

 

[myimmortalfire]

solve equation for x.

ln(y - 1) - ln(2) = x + lnx

my work...
ln((y-l)/2) = x (1 + ln1)

ln((y -1)/2) = x (l +0)

therefore x is ln((y -1)/2)

is this correct?

 

[stapel]

solve equation for x.

ln(y - 1) - ln(2) = x + lnx

my work...
ln((y-l)/2) = x (1 + ln1) <= NOOOoooooooo!!

The logarithm is a function, not a variable. Just as "f(x)" is not "f multiplied by x", so also "ln(x)" is not "l times n times x". There is no "factoring out" that you can with this!! :shock:

Has your class covered functions and function notation yet? If so, has your class covered logarithms yet? (We can find lessons for you, but narrowing down the search parameters would be helpful.)

Thank you!

Eliz.

 

[myimmortalfire]

solve equation for x.

ln(y - 1) - ln(2) = x + lnx

my work...
ln((y-l)/2) - lnx = x

so, would x = ln((y-1)/2x) or will i obtain an answer that will be in decimal form?

 

[myimmortalfire]

question

where will i find more information on parameters

thx
nsn

 

[stapel]

solve equation for x.

Note: This equation cannot actually be solved (by algebraic means, anyway) for "x=". Sorry.

where will i find more information on parameters

"More" information...? :?:

What sort of "parameter" situation are you talking about? Please be specific. Thank you!

Eliz.

 

[morson]

3)

\(\displaystyle ln(y - 1) - ln(2) = x + ln(x)\)

\(\displaystyle ln(\frac{y - 1}{2}\) = ln(e^x) + ln(x)\)

\(\displaystyle ln(\frac{y - 1}{2}\) = ln(xe^x)\)

So: \(\displaystyle \frac{y - 1}{2}\ = xe^x\)

In fact, this does have a solution for x:

\(\displaystyle \L\ x = W(\frac{y - 1}{2}\)\)

Where W(z) is the Lambert W function. A perfectly acceptable solution. The others can be solved easily. Look up quadratic equations.