Evaluating An Exponential Function

Given that, $f(x)=\frac{e^x+e^-x}{e^x-e^-x}$ . Then, we have -------

$\large{f(x)=\frac{e^x+\frac{1}{e^x}}{e^x-\frac{1}{e^x}}}$

$\large{\implies f(x)=\frac{\frac{e^{2x}+1}{e^x}}{\frac{e^{2x}-1}{e^x}}}$

$\large{\implies f(x)=\frac{e^{2x}+1}{e^{2x}-1}}$ .....(i)

From (i), we get----

$\large{\frac{e^{2a}+1}{e^{2a}-1}=f(a)=\frac{5}{3}}$

$\large{3e^{2a}+3=5e^{2a}-5 \implies 2e^{2a}=8 \implies e^{2a}=4}$ .....(ii)

Also, we get from (i)-----

$\large{\frac{e^{2b}+1}{e^{2b}-1}=f(b)=\frac{7}{5}}$

$\large{5e^{2b}+5=7e^{2b}-7 \implies 2e^{2b}=12 \implies e^{2b}=6}$ .....(iii)

Then, from (i),(ii) and (iii), we have----

$\large{f(a+b)=\frac{e^{2(a+b)}+1}{e^{2(a+b)}-1}}=\frac{e^{2a}\times e^{2b}+1}{e^{2a}\times e^{2b}-1}=\frac{4\times 6 +1}{4\times 6 -1}=\frac{24+1}{24-1}=\boxed{\frac{25}{23}}$

simply e^a is 4 and e^-a is 1 by comparing..and e^b and e^-b is 6 and 1 so for e^a+b simply multiply..game over..real easy

using first equality you can see easily that $e^{2a}$ is 4 and from second, you can see that $e^{2b}$ is 6. Now, $f(a+b)$ is simply $(e^{2a} \times e^{2b} + 1) /(e^{2a} \times e^{2b} - 1)$ . Just replace values!

At first solve f(a)=5/3 and f(b)=7/5 to get e^{a} and e^{b} respectively. Now substitute these obtained values in f(a+b) to get it's value.

If you know hyperbolic functions, you can use them to make this problem easier; we can see that $\dfrac{e^x+e^{-x}}{e^x-e^{-x}} = \coth{(x)}$ Using hyperbolic identity, we get $\coth{(a+b)} = \dfrac{1+\coth{(a)}\coth{(b)}}{\coth{(a)}+\coth{(b)}} = \dfrac{1+f(a)f(b)}{f(a)+f(b)}$ Substitute values to get answer. $\dfrac{1+\left(\dfrac{5}{3}\right)\left(\dfrac{7}{5}\right)}{\dfrac{5}{3}+\dfrac{7}{5}} = \boxed{\dfrac{25}{23}}$

5/3=(4+1)/(4-1) ... 7/5=(6+1)/(6-1) -> from answers: x/23=(z+1)/(z-1) -> z=24, -> x=25 ... 25/23

Get the value of 2a from f(a)=5/3 which comes to be log4 and in the same way get the value of 2b from f(b)=7/5 which will comes to be log6.

Hence the value of 2a+2b=log24

Now put the value of 2(a+b)=log24 in f(a+b) and get the answer 25/23.

It's easy.

e^{2a} = 4 e^{2b} = 6

Solving the equation answer is 25/23

$\frac{e^{a}+e^{-a}}{e^{a}-e^{-a}}=\frac{5}{3}$

We suppose e^{a}+e^{-a} = 5 and e^{a}-e^{-a} = 3 <=> e^{a} = 4 and e^{-a} = 1.

So on, we have e^{b} = 6 and e^{-b} = 1.

$\frac{e^{a+b}+e^{-a-b}}{e^{a+b}-e^{-a-b}}=\frac{e^{a}e^{b}+e^{-a}e^{-b}}{e^{a}e^{b}-e^{-a}e^{-b}}=\frac{4*6+1}{4*6-1}=\frac{25}{23}$

$f(a)=\frac{e^a+e^{-a}}{e^a-e^{-a}}=\frac{5}{3}$ $\frac{e^a+e^{-a}+e^a-e^{-a}}{e^a+e^{-a}-e^a-e^{-a}}=\frac{5+3}{5-3}$ $\frac{e^a}{e^(-a)}=\frac{8}{2}$ $e^{2a}=4$

$f(b)=\frac{e^b+e^{-b}}{e^b-e^{-b}}=\frac{7}{5}$ $\frac{e^b+e^{-b}+e^b-e^{-b}}{e^b+e^{-b}-e^b-e^{-b}}=\frac{7+5}{7-5}$ $\frac{e^b}{e^(-b)}=\frac{12}{2}$ $e^{2b}=6$

$f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}$ $f(x)=\frac{e^{-x}(1+e^{2x})}{e^{-x}(e^{2x}-1)}$ $f(x)=\frac{1+e^{2x}}{e^{2x}-1}$ $f(a+b)=\frac{1+e^{2(a+b)}}{e^{2(a+b)}-1}$ $f(a+b)=\frac{1+e^{2a}*e^{2b}} {e^{2a}*e^{2b}-1}=\frac{1+6*4}{6*4-1}=\frac{25}{23}$

upon solving for a and b for the given values, we get- a = (log4)/2 and b = (log6)/2 solve substituting these values in the equation of f(x)