# how can i solve this equation for x

• Nov 29th 2009, 12:52 PM
twostep08
how can i solve this equation for x
this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

3/4=(e^x - e^-x)/(e^x + e^-x)

any help is very appreciated
• Nov 29th 2009, 12:54 PM
e^(i*pi)
Quote:

Originally Posted by twostep08
this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

3/4=(e^x - e^-x)/(e^x + e^-x)

any help is very appreciated

You can just run $\displaystyle arctanh \left(\frac{3}{4}\right)$

my calculator says it is equal to 0.973 (3sf)
• Nov 29th 2009, 12:57 PM
twostep08
im not really allowed to use my calculator...there must be some nice answer using logs?
• Nov 29th 2009, 01:15 PM
e^(i*pi)
Quote:

Originally Posted by twostep08
this is part of a larger question for hyperbolic functions, but i feel if i get this the rest will come into place...what is x of tanh x= 3/4, or you can solve the below equation

3/4=(e^x - e^-x)/(e^x + e^-x)

any help is very appreciated

Yeah there is. You can use standard algebraic techniques and logs. Ignore the second solution if you only want a real answer

$\displaystyle 3e^x+3e^{-x} = 4e^x-4e^{-x}$

$\displaystyle e^x-7e^{-x} = 0$

$\displaystyle e^x - \frac{7}{e^x} = 0$

$\displaystyle e^{2x} - 7 = 0$

$\displaystyle (e^x-\sqrt{7})(e^x+\sqrt{7}) = 0$

$\displaystyle x = \frac{1}{2}ln(7)$ or $\displaystyle x = \frac{1}{2}ln(-7) = \frac{1}{2}ln(7)+ i \left(\frac{\pi}{2}\right)$
• Nov 29th 2009, 01:31 PM
twostep08
ok im following you all the way to the last step when you say e^x - (sq.rt.)7 = 1/2ln(7)

what makes that so?
• Nov 29th 2009, 01:36 PM
e^(i*pi)
Quote:

Originally Posted by twostep08
ok im following you all the way to the last step when you say e^x - (sq.rt.)7 = 1/2ln(7)

what makes that so?

It copmes from the difference of two squares when factorising the quadratic and the term is equal to 0

$\displaystyle e^x-\sqrt7 = 0$

From here add $\displaystyle \sqrt7$ to both sides and take the natural log

I used the log power law to bring the 1/2 to the front
• Nov 29th 2009, 01:37 PM
Defunkt
$\displaystyle e^x-\sqrt{7} = 0 \Rightarrow log(e^x) = x = log(\sqrt{7}) = \frac{1}{2}log7$
• Nov 29th 2009, 01:43 PM
e^(i*pi)
Quote:

Originally Posted by Defunkt
$\displaystyle e^x-\sqrt{7} = 0 \Rightarrow log(e^x) = x = log(\sqrt{7}) = \frac{1}{2}log7$

Just to point out that Defunkt is also referring to the natural log ie base e

$\displaystyle log_e(e^x) = ln(e^x) = x$