Solve Lt = (2 ^ y - 2 ^ (- y))/(2 ^ y + 2 ^ (- y)) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$Lt=\frac{2^{y}-2^{-y}}{2^{y}+2^{-y}}$$

Solve for L

$\left\{\begin{matrix}L=-\frac{1-4^{y}}{t\left(4^{y}+1\right)}\text{, }&t\neq 0\\L\in \mathrm{R}\text{, }&y=0\text{ and }t=0\end{matrix}\right.$

Steps for Solving Linear Equation

Factor the expressions that are not already factored in $\frac{2^{y}-2^{-y}}{2^{y}+2^{-y}}$.

$$Lt=\frac{\frac{1}{2^{y}}\left(2^{y}-1\right)\left(2^{y}+1\right)}{\frac{1}{2^{y}}\left(\left(2^{y}\right)^{2}+1\right)}$$

Cancel out $\frac{1}{2^{y}}$ in both numerator and denominator.

$$Lt=\frac{\left(2^{y}-1\right)\left(2^{y}+1\right)}{\left(2^{y}\right)^{2}+1}$$

Consider $\left(2^{y}-1\right)\left(2^{y}+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$Lt=\frac{\left(2^{y}\right)^{2}-1}{\left(2^{y}\right)^{2}+1}$$

The equation is in standard form.

$$tL=\frac{4^{y}-1}{4^{y}+1}$$

Divide both sides by $t$.

$$\frac{tL}{t}=\frac{4^{y}-1}{\left(4^{y}+1\right)t}$$

Dividing by $t$ undoes the multiplication by $t$.

$$L=\frac{4^{y}-1}{\left(4^{y}+1\right)t}$$

Divide $\frac{4^{y}-1}{4^{y}+1}$ by $t$.

$$L=\frac{4^{y}-1}{t\left(4^{y}+1\right)}$$

Solve for t

$\left\{\begin{matrix}t=-\frac{1-4^{y}}{L\left(4^{y}+1\right)}\text{, }&L\neq 0\\t\in \mathrm{R}\text{, }&y=0\text{ and }L=0\end{matrix}\right.$

Steps for Solving Linear Equation

Factor the expressions that are not already factored in $\frac{2^{y}-2^{-y}}{2^{y}+2^{-y}}$.

$$Lt=\frac{\frac{1}{2^{y}}\left(2^{y}-1\right)\left(2^{y}+1\right)}{\frac{1}{2^{y}}\left(\left(2^{y}\right)^{2}+1\right)}$$

Cancel out $\frac{1}{2^{y}}$ in both numerator and denominator.

$$Lt=\frac{\left(2^{y}-1\right)\left(2^{y}+1\right)}{\left(2^{y}\right)^{2}+1}$$

Consider $\left(2^{y}-1\right)\left(2^{y}+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$Lt=\frac{\left(2^{y}\right)^{2}-1}{\left(2^{y}\right)^{2}+1}$$

The equation is in standard form.

$$Lt=\frac{4^{y}-1}{4^{y}+1}$$

Divide both sides by $L$.

$$\frac{Lt}{L}=\frac{4^{y}-1}{\left(4^{y}+1\right)L}$$

Dividing by $L$ undoes the multiplication by $L$.

$$t=\frac{4^{y}-1}{\left(4^{y}+1\right)L}$$

Divide $\frac{4^{y}-1}{4^{y}+1}$ by $L$.

$$t=\frac{4^{y}-1}{L\left(4^{y}+1\right)}$$