Converting exponentials to base e

As mentioned in the last topic, the gradient of e^x is exactly the same as the y value at that point.

That makes a lot of things much easier for us to work with!

We can actually use this power of base e exponentials with any exponential function - by converting them to base e.

a=e^{\ln(a)}

The key to converting any exponential function to base e is the fact that we can rewrite any number a as e^{\ln(a)}.

Basically, what we’re doing here is finding what number we need to raise e to in order to get a - and that number is \ln(a).

Then, we raise e to that power to get a back.

While it might seem like we’ve complicated things, we now have a term in base e, which we can find the gradient of!

Converting a^x to base e

Knowing that a=e^{\ln(a)}, we can rewrite a^x as (e^{\ln(a)})^x.

Then, we can use the rule of indices that says that (a^b)^c = a^{bc} to rewrite (e^{\ln(a)})^x as e^{x\ln(a)}.

Finding the gradient of a^x

Now we have a^x rewritten as e^{x\ln(a)}, we can find the gradient of it using the rule that \frac{d}{dx}e^{f(x)} = f'(x)e^{f(x)} (see base e exponentials for more on that rule).

In this case, f(x) = x\ln(a), so f'(x) = \ln(a) (because \ln(a) is a constant, so it just comes out of the differentiation).

Then, we can apply the rule to get \frac{d}{dx}e^{x\ln(a)} = \ln(a)e^{x\ln(a)}.

Finally, we can rewrite e^{x\ln(a)} back to a^x to get the final answer of \frac{d}{dx}a^x = \ln(a)a^x.

\frac{d}{dx}a^x = \ln(a)a^x for any a > 0 and a \neq 1.

Example: 2^x

Convert 2^x to base e

• 2^x = (e^{\ln(2)})^x (because 2 = e^{\ln(2)})
• 2^x = e^{x\ln(2)} (using the rule of indices)
• Answer: e^{x\ln(2)}

Find the gradient of 2^x

• \frac{d}{dx}2^x = \ln(2)2^x (using the rule for finding the gradient of a^x)
• Answer: \ln(2)2^x

Find the gradient of 2^x at the point (1, 2)

• \frac{d}{dx}2^x = \ln(2)2^x (using the rule for finding the gradient of a^x)
• \frac{d}{dx}2^x = \ln(2)2^1 (substituting x=1 into the expression for the gradient)
• \frac{d}{dx}2^x = \ln(2) \cdot 2 (simplifying)
• Answer: \ln(2) \cdot 2

Example: 5^2x

Convert 5^{2x} to base e

• 5^{2x} = (e^{\ln(5)})^{2x} (because 5 = e^{\ln(5)})
• 5^{2x} = e^{2x\ln(5)} (using the rule of indices)
• Answer: e^{2x\ln(5)}

Find the gradient of 5^{2x}

• \frac{d}{dx}5^{2x} = 2\ln(5)5^{2x} (using the rule for finding the gradient of a^x)
• Answer: 2\ln(5)5^{2x}

Find the gradient of 5^{2x} at the point (0, 1)

• \frac{d}{dx}5^{2x} = 2\ln(5)5^{2x} (using the rule for finding the gradient of a^x)
• \frac{d}{dx}5^{2x} = 2\ln(5)5^0 (substituting x=0 into the expression for the gradient)
• \frac{d}{dx}5^{2x} = 2\ln(5) \cdot 1 (simplifying)
• Answer: 2\ln(5)

Find the gradient of 5^{2x} at the point (1, 25)

• \frac{d}{dx}5^{2x} = 2\ln(5)5^{2x} (using the rule for finding the gradient of a^x)
• \frac{d}{dx}5^{2x} = 2\ln(5)5^2 (substituting x=1 into the expression for the gradient)
• \frac{d}{dx}5^{2x} = 2\ln(5) \cdot 25 (simplifying)
• Answer: 50\ln(5)

Example: 10^{-x}

Convert 10^{-x} to base e

• 10^{-x} = (e^{\ln(10)})^{-x} (because 10 = e^{\ln(10)})
• 10^{-x} = e^{-x\ln(10)} (using the rule of indices)
• Answer: e^{-x\ln(10)}

Find the gradient of 10^{-x}

• \frac{d}{dx}10^{-x} = -\ln(10)10^{-x} (using the rule for finding the gradient of a^x)
• Answer: -\ln(10)10^{-x}

Find the gradient of 10^{-x} at the point (0, 1)

• \frac{d}{dx}10^{-x} = -\ln(10)10^{-x} (using the rule for finding the gradient of a^x)
• \frac{d}{dx}10^{-x} = -\ln(10)10^0 (substituting x=0 into the expression for the gradient)
• \frac{d}{dx}10^{-x} = -\ln(10) \cdot 1 (simplifying)
• Answer: -\ln(10)

Find the gradient of 10^{-x} at the point (1, 0.1)

• \frac{d}{dx}10^{-x} = -\ln(10)10^{-x} (using the rule for finding the gradient of a^x)
• \frac{d}{dx}10^{-x} = -\ln(10)10^{-1} (substituting x=1 into the expression for the gradient)
• \frac{d}{dx}10^{-x} = -\ln(10) \cdot 0.1 (simplifying)
• Answer: -0.1\ln(10)

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