The Laplace transform of 2^t is equal to 1/(s-ln2), provided that s>ln2, where ln x = log_{e}x. This can be determined by the definition of Laplace transforms as well as by the Laplace transform formula of e^{at}. In this post, we will use both methods to find the Laplace transform of 2 to the power t.

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## Laplace transform of 2^{t }Formula

The Laplace transform formula of 2^{t} (2 to the power t) is given below:

L{2^{t}} = 1/(s-ln 2),

where s>ln2 and ln denotes the natural logarithm (that is, the logarithm with base e).

Next, we will learn how to find the Laplace transform of 2^{t}.

## Find Laplace transform of 2^{t }

**Method 1** (using the Laplace transform formula of e^{at}):

We know that

L{e^{at}} = 1/(s-a), provided that s>a.

As 2^{t} can be written as 2^{t}= (e^{ln2})^{t}, applying the above formula, we get that

L{2^{t}} = 1/(s-ln2) , provided that s>ln2.

**Method 2** (using definition):

## Laplace transform of 2^{t }by Definition

Recall, the definition of the Laplace transform of f(t), denoted by F(s) or L{f(t)}, is given as follows:

F(s) = L{f(t)} = $\int_0^\infty$ e^{-st} f(t) dt.

So the Laplace transform of 2 to the power t by definition will be

F(s) = L{2^{t}} = $\int_0^\infty$ e^{-st} 2^{t} dt.

Use the fact 2^{t} = (e^{ln2})^{t}. So

L{2^{t}} = $\int_0^\infty$ e^{-st} e^{tln2} dt

= $\int_0^\infty$ e^{t(ln2-s)} dt

= $\left[ \dfrac{e^{t(\ln 2-s)}}{\ln 2-s} \right]_0^\infty$

= $\dfrac{1}{\ln 2-s}$ $\left( \lim\limits_{t \to \infty} e^{t(\ln 2-s)} – e^{0(\ln2 -s)}\right)$ **…(I)**

= $\dfrac{1}{\ln 2-s}(0-1)$

= $\dfrac{1}{s-\ln 2}$.

Note that the above limit in **(I)** exists only when s>ln 2; otherwise, that is when s≤ln 2 the limit does not exist. Thus, we obtain:

The Laplace transform of 2^{t} is 1/(s-ln2) and it exists when s>ln2. This is proved by the definition of Laplace transforms.

**Have You Read These Laplace Transforms:**

## FAQs

**Q1: What is Laplace transform of 2^t?**

Answer: The Laplace transform of 2^t is given by L{2^{t}} = 1/(s-log_{e}2), where s>log_{e}2.