In mathematics, many logarithmic identities exist. The following is a compilation of the notable of these, many of which are used for computational purposes.
Trivial identities Trivial mathematical identities are relatively simple (for an experienced mathematician), though not necessarily unimportant. The trivial logarithmic identities are as follows:
Explanations By definition, we know that:
log b ( y ) = x ⟺ b x = y , {\displaystyle \log _{b}(y)=x\iff b^{x}=y,}
where b ≠ 0 {\displaystyle b\neq 0} and b ≠ 1 {\displaystyle b\neq 1} . Setting x = 0 {\displaystyle x=0} , we can see that:
b x = y ⟺ b ( 0 ) = y ⟺ 1 = y ⟺ y = 1 {\displaystyle b^{x}=y\iff b^{(0)}=y\iff 1=y\iff y=1}
So, substituting these values into the formula, we see that:
log b ( y ) = x ⟺ log b ( 1 ) = 0 , {\displaystyle \log _{b}(y)=x\iff \log _{b}(1)=0,}
which gets us the first property. Setting x = 1 {\displaystyle x=1} , we can see that:
b x = y ⟺ b ( 1 ) = y ⟺ b = y ⟺ y = b {\displaystyle b^{x}=y\iff b^{(1)}=y\iff b=y\iff y=b}
So, substituting these values into the formula, we see that:
log b ( y ) = x ⟺ log b ( b ) = 1 , {\displaystyle \log _{b}(y)=x\iff \log _{b}(b)=1,}
which gets us the second property.
Cancelling exponentials Logarithms and exponentials with the same base cancel each other. This is true because logarithms and exponentials are inverse operations – much like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations:
b log b ( x ) = x because antilog b ( log b ( x ) ) = x {\displaystyle b^{\log _{b}(x)}=x{\text{ because }}{\mbox{antilog}}_{b}(\log _{b}(x))=x}
log b ( b x ) = x because log b ( antilog b ( x ) ) = x {\displaystyle \log _{b}(b^{x})=x{\text{ because }}\log _{b}({\mbox{antilog}}_{b}(x))=x}
Both of the above are derived from the following two equations that define a logarithm: (note that in this explanation, the variables of x {\displaystyle x} and x {\displaystyle x} may not be referring to the same number)
log b ( y ) = x ⟺ b x = y {\displaystyle \log _{b}(y)=x\iff b^{x}=y}
Looking at the equation b x = y {\displaystyle b^{x}=y} , and substituting the value for x {\displaystyle x} of
log b ( y ) = x {\displaystyle \log _{b}(y)=x} , we get the following equation:
b x = y ⟺ b log b ( y ) = y ⟺ b log b ( y ) = y , {\displaystyle b^{x}=y\iff b^{\log _{b}(y)}=y\iff b^{\log _{b}(y)}=y,}
which gets us the first equation. Another more rough way to think about it is that b something = y {\displaystyle b^{\text{something}}=y} , and that that " something {\displaystyle {\text{something}}} " is log b ( y ) {\displaystyle \log _{b}(y)} . Looking at the equation log b ( y ) = x {\displaystyle \log _{b}(y)=x} , and substituting the value for y {\displaystyle y} of b x = y {\displaystyle b^{x}=y} , we get the following equation:
log b ( y ) = x ⟺ log b ( b x ) = x ⟺ log b ( b x ) = x , {\displaystyle \log _{b}(y)=x\iff \log _{b}(b^{x})=x\iff \log _{b}(b^{x})=x,}
which gets us the second equation. Another more rough way to think about it is that log b ( something ) = x {\displaystyle \log _{b}({\text{something}})=x} , and that that " something {\displaystyle {\text{something}}} " is b x {\displaystyle b^{x}} .
Using simpler operations Logarithms can be used to make calculations easier. For example, two numbers can be multiplied just by using a logarithm table and adding. These are often known as logarithmic properties, which are documented in the table below. The first three operations below assume that x = bc and/or y = bd, so that logb(x) = c and logb(y) = d. Derivations also use the log definitions x = blogb(x) and x = logb(bx).
Where b {\displaystyle b} , x {\displaystyle x} , and y {\displaystyle y} are positive real numbers and b ≠ 1 {\displaystyle b\neq 1} , and c {\displaystyle c} and d {\displaystyle d} are real numbers. The laws result from canceling exponentials and the appropriate law of indices. Starting with the first law:
x y = b log b ( x ) b log b ( y ) = b log b ( x ) + log b ( y ) ⇒ log b ( x y ) = log b ( b log b ( x ) + log b ( y ) ) = log b ( x ) + log b ( y ) {\displaystyle xy=b^{\log _{b}(x)}b^{\log _{b}(y)}=b^{\log _{b}(x)+\log _{b}(y)}\Rightarrow \log _{b}(xy)=\log _{b}(b^{\log _{b}(x)+\log _{b}(y)})=\log _{b}(x)+\log _{b}(y)}
The law for powers exploits another of the laws of indices:
x y = ( b log b ( x ) ) y = b y log b ( x ) ⇒ log b ( x y ) = y log b ( x ) {\displaystyle x^{y}=(b^{\log _{b}(x)})^{y}=b^{y\log _{b}(x)}\Rightarrow \log _{b}(x^{y})=y\log _{b}(x)}
The law relating to quotients then follows:
log b ( x y ) = log b ( x y − 1 ) = log b ( x ) + log b ( y − 1 ) = log b ( x ) − log b ( y ) {\displaystyle \log _{b}{\bigg (}{\frac {x}{y}}{\bigg )}=\log _{b}(xy^{-1})=\log _{b}(x)+\log _{b}(y^{-1})=\log _{b}(x)-\log _{b}(y)}
log b ( 1 y ) = log b ( y − 1 ) = − log b ( y ) {\displaystyle \log _{b}{\bigg (}{\frac {1}{y}}{\bigg )}=\log _{b}(y^{-1})=-\log _{b}(y)}
Similarly, the root law is derived by rewriting the root as a reciprocal power:
log b ( x y ) = log b ( x 1 y ) = 1 y log b ( x ) {\displaystyle \log _{b}({\sqrt[{y}]{x}})=\log _{b}(x^{\frac {1}{y}})={\frac {1}{y}}\log _{b}(x)}
Derivations of product, quotient, and power rules These are the three main logarithm laws, rules, or principles, from which the other properties listed above can be proven. Each of these logarithm properties correspond to their respective exponent law, and their derivations and proofs will hinge on those facts. There are multiple ways to derive or prove each logarithm law – this is just one possible method.
Logarithm of a product To state the logarithm of a product law formally:
∀ b ∈ R + , b ≠ 1 , ∀ x , y , ∈ R + , log b ( x y ) = log b ( x ) + log b ( y ) {\displaystyle \forall b\in \mathbb {R} _{+},b\neq 1,\forall x,y,\in \mathbb {R} _{+},\log _{b}(xy)=\log _{b}(x)+\log _{b}(y)}
Derivation: Let b ∈ R + {\displaystyle b\in \mathbb {R} _{+}} , where b ≠ 1 {\displaystyle b\neq 1} , and let x , y ∈ R + {\displaystyle x,y\in \mathbb {R} _{+}} . We want to relate the expressions log b ( x ) {\displaystyle \log _{b}(x)} and log b ( y ) {\displaystyle \log _{b}(y)} . This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to log b ( x ) {\displaystyle \log _{b}(x)} and log b ( y ) {\displaystyle \log _{b}(y)} quite often, we will give them some variable names to make working with them easier: Let m = log b ( x ) {\displaystyle m=\log _{b}(x)} , and let n = log b ( y ) {\displaystyle n=\log _{b}(y)} . Rewriting these as exponentials, we see that
m = log b ( x ) ⟺ b m = x , n = log b ( y ) ⟺ b n = y . {\displaystyle {\begin{aligned}m&=\log _{b}(x)\iff b^{m}=x,\\n&=\log _{b}(y)\iff b^{n}=y.\end{aligned}}}
From here, we can relate b m {\displaystyle b^{m}} (i.e. x {\displaystyle x} ) and b n {\displaystyle b^{n}} (i.e. y {\displaystyle y} ) using exponent laws as
x y = ( b m ) ( b n ) = b m ⋅ b n = b m + n {\displaystyle xy=(b^{m})(b^{n})=b^{m}\cdot b^{n}=b^{m+n}}
To recover the logarithms, we apply log b {\displaystyle \log _{b}} to both sides of the equality.
log b ( x y ) = log b ( b m + n ) {\displaystyle \log _{b}(xy)=\log _{b}(b^{m+n})}
The right side may be simplified using one of the logarithm properties from before: we know that log b ( b m + n ) = m + n {\displaystyle \log _{b}(b^{m+n})=m+n} , giving
log b ( x y ) = m + n {\displaystyle \log _{b}(xy)=m+n}
We now resubstitute the values for m {\displaystyle m} and n {\displaystyle n} into our equation, so our final expression is only in terms of x {\displaystyle x} , y {\displaystyle y} , and b {\displaystyle b} .
log b ( x y ) = log b ( x ) + log b ( y ) {\displaystyle \log _{b}(xy)=\log _{b}(x)+\log _{b}(y)}
This completes the derivation.
Logarithm of a quotient To state the logarithm of a quotient law formally:
∀ b ∈ R + , b ≠ 1 , ∀ x , y , ∈ R + , log b ( x y ) = log b ( x ) − log b ( y ) {\displaystyle \forall b\in \mathbb {R} _{+},b\neq 1,\forall x,y,\in \mathbb {R} _{+},\log _{b}\left({\frac {x}{y}}\right)=\log _{b}(x)-\log _{b}(y)}
Derivation: Let b ∈ R + {\displaystyle b\in \mathbb {R} _{+}} , where b ≠ 1 {\displaystyle b\neq 1} , and let x , y ∈ R + {\displaystyle x,y\in \mathbb {R} _{+}} . We want to relate the expressions log b ( x ) {\displaystyle \log _{b}(x)} and log b ( y ) {\displaystyle \log _{b}(y)} . This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to log b ( x ) {\displaystyle \log _{b}(x)} and log b ( y ) {\displaystyle \log _{b}(y)} quite often, we will give them some variable names to make working with them easier: Let m = log b ( x ) {\displaystyle m=\log _{b}(x)} , and let n = log b ( y ) {\displaystyle n=\log _{b}(y)} . Rewriting these as exponentials, we see that:
m = log b ( x ) ⟺ b m = x , n = log b ( y ) ⟺ b n = y . {\displaystyle {\begin{aligned}m&=\log _{b}(x)\iff b^{m}=x,\\n&=\log _{b}(y)\iff b^{n}=y.\end{aligned}}}
From here, we can relate b m {\displaystyle b^{m}} (i.e. x {\displaystyle x} ) and b n {\displaystyle b^{n}} (i.e. y {\displaystyle y} ) using exponent laws as
x y = ( b m ) ( b n ) = b m b n = b m − n {\displaystyle {\frac {x}{y}}={\frac {(b^{m})}{(b^{n})}}={\frac {b^{m}}{b^{n}}}=b^{m-n}}
To recover the logarithms, we apply log b {\displaystyle \log _{b}} to both sides of the equality.
log b ( x y ) = log b ( b m − n ) {\displaystyle \log _{b}\left({\frac {x}{y}}\right)=\log _{b}\left(b^{m-n}\right)}
The right side may be simplified using one of the logarithm properties from before: we know that log b ( b m − n ) = m − n {\displaystyle \log _{b}(b^{m-n})=m-n} , giving
log b ( x y ) = m − n {\displaystyle \log _{b}\left({\frac {x}{y}}\right)=m-n}
We now resubstitute the values for m {\displaystyle m} and n {\displaystyle n} into our equation, so our final expression is only in terms of x {\displaystyle x} , y {\displaystyle y} , and b {\displaystyle b} .
log b ( x y ) = log b ( x ) − log b ( y ) {\displaystyle \log _{b}\left({\frac {x}{y}}\right)=\log _{b}(x)-\log _{b}(y)}
This completes the derivation.
Logarithm of a power To state the logarithm of a power law formally:
∀ b ∈ R + , b ≠ 1 , ∀ x ∈ R + , ∀ r ∈ R , log b ( x r ) = r log b ( x ) {\displaystyle \forall b\in \mathbb {R} _{+},b\neq 1,\forall x\in \mathbb {R} _{+},\forall r\in \mathbb {R} ,\log _{b}(x^{r})=r\log _{b}(x)}
Derivation: Let b ∈ R + {\displaystyle b\in \mathbb {R} _{+}} , where b ≠ 1 {\displaystyle b\neq 1} , let x ∈ R + {\displaystyle x\in \mathbb {R} _{+}} , and let r ∈ R {\displaystyle r\in \mathbb {R} } . For this derivation, we want to simplify the expression log b ( x r ) {\displaystyle \log _{b}(x^{r})} . To do this, we begin with the simpler expression log b ( x ) {\displaystyle \log _{b}(x)} . Since we will be using log b ( x ) {\displaystyle \log _{b}(x)} often, we will define it as a new variable: Let m = log b ( x ) {\displaystyle m=\log _{b}(x)} . To more easily manipulate the expression, we rewrite it as an exponential. By definition, m = log b ( x ) ⟺ b m = x {\displaystyle m=\log _{b}(x)\iff b^{m}=x} , so we have
b m = x {\displaystyle b^{m}=x}
Similar to the derivations above, we take advantage of another exponent law. In order to have x r {\displaystyle x^{r}} in our final expression, we raise both sides of the equality to the power of r {\displaystyle r} :
( b m ) r = ( x ) r b m r = x r {\displaystyle {\begin{aligned}(b^{m})^{r}&=(x)^{r}\\b^{mr}&=x^{r}\end{aligned}}}
where we used the exponent law ( b m ) r = b m r {\displaystyle (b^{m})^{r}=b^{mr}} . To recover the logarithms, we apply log b {\displaystyle \log _{b}} to both sides of the equality.
log b ( b m r ) = log b ( x r ) {\displaystyle \log _{b}(b^{mr})=\log _{b}(x^{r})}
The left side of the equality can be simplified using a logarithm law, which states that log b ( b m r ) = m r {\displaystyle \log _{b}(b^{mr})=mr} .
m r = log b ( x r ) {\displaystyle mr=\log _{b}(x^{r})}
Substituting in the original value for m {\displaystyle m} , rearranging, and simplifying gives
( log b ( x ) ) r = log b ( x r ) r log b ( x ) = log b ( x r ) log b ( x r ) = r log b ( x ) {\displaystyle {\begin{aligned}\left(\log _{b}(x)\right)r&=\log _{b}(x^{r})\\r\log _{b}(x)&=\log _{b}(x^{r})\\\log _{b}(x^{r})&=r\log _{b}(x)\end{aligned}}}
This completes the derivation.
Changing the base Most calculators have buttons for natural logarithms (ln) and common logarithms (log or log10), but not all calculators have buttons for the logarithm of an arbitrary base. Accordingly, it is sometimes useful to change the base of a logarithm.As briefly mentioned in the § Trivial identities section, the definition of a logarithm is: log b