Natural logarithms are logarithms to the base of e (Euler's number = 2.71828 ...). They are expressed as \(\log_e(x)\) and they can be written as \(\ln (x)\) as shorthand.
Converting between natural logarithms and exponential functions
Start with a general natural logarithm: \(\ln(x) = y\). You can easily rewrite this as \(\log_e(x) = y\).
As always, you need to label each part of the function: the base is e (as with natural logarithms), the exponent is y, and the answer of the exponential is x.
Therefore you can rewrite logarithms as \(e^y = x\).
Solve \(e^x = 5\) to 3 sf
- Label each part of the function so: the base is e, the exponent is x, and the answer of the exponential is 5.
- Then formulate it into a logarithm to become \(\log_e(5) = x\)
- As Loge is the same as saying Ln, you can write it as a natural logarithm
\(\ln(5) = x = 1.61 (3 s.f)\)
What are the rules for natural logarithms?
Along with the specific rules for natural logarithms, you can use the general Laws of Logs as well as the Exponential Rules.
Rules for natural logarithms
- \(\ln(e) = 1\)
- \(\ln(1) = 0\)
- \(\ln(e^x) = x\)
- If \(\ln(y) = \ln(x)\), then y = x
- \(e^{\ln(x)} = x\)
Proving natural logarithm rules
Just like the proofs for Laws of Logs, you need to be able to understand each step of proving a natural logarithm rule – you do not need to feel like you could have got to that point without any help.
Proving Ln (1) = 0
\(\ln(1) = m\) can be written as \(\log_e(1) = m\)
You will rewrite it as an exponential function where the base is e, the answer of the exponential is 1, and the exponent is m. This exponential would look like this: \(e^m = 1\)
Using our Power = 0 exponential law, you know that the exponent (in this case, m) must be 0 for the answer to the exponential to be 1.
Thus \(\ln(1) = 0\)
Proving Ln (e) = 1
\(\ln(e) = n\) can be rewritten as \(\log_e(e) = n\) where the base is e, the answer to the exponential is e, and the exponent is n.
As a result, you rewrite \(\log_e(e) = n\) as \(e^n = e\).
According to our exponential rules, when the answer to the exponential is the same as the base, then the power must be 1.
Thus, \(\ln(e) = 1\)
Proving Ln(ex)=x
As exponential and logarithms are inverse functions, they cancel each other out when they are placed in the same function.
This concept is the same as multiplying a number by 2 and then dividing by 2 – you end up with the same number you have in the beginning.
Therefore, ln and e will cancel out so that you are left with just x.
Proving Ln (y) = Ln (x) means y = x
If you set ln(y) = a and ln(x) = b, you can rewrite each function as an exponential.
- \(\ln(y) = a, \log_e(y) = a\)
Where the base is e, the exponent is a, and answer to the exponential is y. Therefore, the exponential is \(e^a = y\).
- \(\ln(x) = b, \log_e(x) = b\)
Where the base is e, the exponent is b, and the answer to the exponential is x. Therefore, the exponential is \(e^b = x\).
Because you are told ln (y) = ln (x), \(e^a\) must be equal to \(e^b\), therefore y = x.
Proving eLn(x)=x
This law uses the same thinking as the \(\ln(e^x) = x\)
Application of natural logarithm rules
Example 1: Solve \(e^{2x} = 6\)
The expression \(e^{2x} = 6\) can be written as a natural logarithm as the base is e, the exponent is 2x, and the answer to the exponential is 6.
So as a natural logarithm, it could be written as ln (6) = 2x.
Therefore, \(\frac{\ln(6)}{2} = 0.896 (3 s.f)\)
Example 2: Solve \(e^{x+3} = 10\)
The expression \(e^{x+3}\) can be written as a logarithm, whereby the base is e; the exponent is x + 3, and the answer to the exponential is 10.
\(\ln(10) = x + 3\)
Therefore, \(x = \ln(10) - 3 = -0.697(3 s.f)\)
Example 3: Solve \(e^{\ln(x^3)} = 8\)
As the exponential and logarithms are inverse functions, the e and Ln will cancel each other.
Therefore, \(x^3 = 8; x = 2\)
Example 4: Solve \(\ln(x+1) = 1.4\)
To get x on its own, we need to convert the logarithm to an exponential where the base is e, the exponent is 1.4, and the answer to the exponential is x + 1.
Therefore, \(e^{1.4} = x+1\) and \(x = e^{1.4} -1 = 3.06(3 s.f)\)
Example 5: Solve \(2\ln(6) + \ln(2) - \ln(4) = x\)
1. Due to the power logarithm rule, \(2\ln(6)\) can be written as \(\ln(6^2) = \ln(36)\)
Therefore, \(\ln(36) +\ln(2) - \ln(4) = x\)
2. Using the product and quotient rule, we can do this further:
\(\ln(36 \cdot 2) - \ln(4) = x\)
\(\ln(\frac{36 \cdot 2}{4}) = x\)
\(\ln(\frac{72}{4}) = \ln(18) = x = 2.89 (3 s.f)\)
Natural Logarithm - Key takeaways
- Natural logarithms are logarithms with the base of e.
- To use natural logarithms to solve and simplify, you can use:\(\ln(1) = 0\); \(\ln(e) = 1\); if \(\ln(y) = \ln(x)\), then y = x; \(e^{\ln(x)} = x\). This is alongside the laws for both Exponentials and Logarithms.
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