# Thread: how can i solve this equation for x

1. ## 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

2. 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)

3. im not really allowed to use my calculator...there must be some nice answer using logs?

4. 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 $\displaystyle x = \frac{1}{2}ln(-7) = \frac{1}{2}ln(7)+ i \left(\frac{\pi}{2}\right)$

5. 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?

6. 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

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

8. 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

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