If
$$f(x)=\frac{2^{2x}+2^{-x}}{2^{x}-2^{-x}}$$
Then evaluate $f(\log_2(3))$.
Can someone help me to understand the calculation?
I figured out that the result is $7/2$ but I have problems solving by hand.
I know that $2\uparrow {2\left(\log_2(3)\right)}$ is $9$ but could someone explain to me how to get to this $9$ and how to calculate negative logarithms as power?
$$f(\log_23):=\frac{2^{2\log_23}+2^{-\log_23}}{2^{\log_23}-2^{-\log_23}}$$
But, as in the comment,
$$2^{\log_2b}=b=\log_22^b\;,\;\;\text{and of course}\;\;\log_ax^n=n\log_ax$$
HINT: $$\frac{2^{2\log_2 3} + 2^{-\log_2 3}}{2^{\log_2 3} - 2^{-\log_2 3}} = \frac{2^{\log_2 3^2} + 2^{\log_2 3^{-1}}}{2^{\log_2 3} - 2^{\log_2 3^{-1}}} = ...$$ and $$2^{\log_2 x} = x$$
Let us first clarify some required basic points about exponents and logarithms that may be of help to you:
$n^{-1} = \frac{1}{n} \implies n^{-m} = \frac{1}{n^m}$
$n^ {xy} = (n^x)^y = (n^y)^x$
$b^{\log_b a} = a$
$$ \require{cancel} \begin{align} f(x) &:= \frac{2^{2x} + 2^{-x}}{2^x - 2^{-x}} \cdot 1\\ &= \frac{(2^{x})^2 + \frac{1}{2^x}}{2^x - \frac{1}{2^x}} \cdot \frac{2^x}{2^x}\\ &= \frac{(2^x)^3 + 1^3}{(2^x)^2 - 1^2}\\ &= \frac{\cancel{(2^x + 1)}\left((2^x)^2 - 2^x + 1\right)}{\cancel{(2^x + 1)}(2^x - 1)}\\ &= \frac{2^{2x} - 2^x + 1}{2^x - 1} \end{align} $$ At $x = \log_2 (3)$ ,
$$ \begin{align} f(\log_2 (3)) &= \frac{(2^{\log_2 (3)})^2 - 2^{\log_2 (3)} + 1}{2^{\log_2 (3)} - 1}\\ &= \frac{3^2 - 3 + 1}{3 -1} \\ &= \frac{7}{2} = 3.5 \end{align} $$