[{"@context":"http:\/\/schema.org\/","@type":"BlogPosting","@id":"https:\/\/wiki.edu.vn\/en\/wiki14\/list-of-logarithmic-identities-wikipedia\/#BlogPosting","mainEntityOfPage":"https:\/\/wiki.edu.vn\/en\/wiki14\/list-of-logarithmic-identities-wikipedia\/","headline":"List of logarithmic identities – Wikipedia","name":"List of logarithmic identities – Wikipedia","description":"before-content-x4 In mathematics, many logarithmic identities exist. The following is a compilation of the notable of these, many of which","datePublished":"2021-12-23","dateModified":"2021-12-23","author":{"@type":"Person","@id":"https:\/\/wiki.edu.vn\/en\/wiki14\/author\/lordneo\/#Person","name":"lordneo","url":"https:\/\/wiki.edu.vn\/en\/wiki14\/author\/lordneo\/","image":{"@type":"ImageObject","@id":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","height":96,"width":96}},"publisher":{"@type":"Organization","name":"Enzyklop\u00e4die","logo":{"@type":"ImageObject","@id":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","url":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","width":600,"height":60}},"image":{"@type":"ImageObject","@id":"https:\/\/wikimedia.org\/api\/rest_v1\/media\/math\/render\/svg\/901f6efd3f7b26aa95b855e884a8c2c620ef1fe0","url":"https:\/\/wikimedia.org\/api\/rest_v1\/media\/math\/render\/svg\/901f6efd3f7b26aa95b855e884a8c2c620ef1fe0","height":"","width":""},"url":"https:\/\/wiki.edu.vn\/en\/wiki14\/list-of-logarithmic-identities-wikipedia\/","wordCount":36367,"articleBody":"     (adsbygoogle = window.adsbygoogle || []).push({});before-content-x4In mathematics, many logarithmic identities exist. The following is a compilation of the notable of these, many of which are used for computational purposes.       (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4Table of ContentsTrivial identities[edit]Explanations[edit]Cancelling exponentials[edit]Using simpler operations[edit]Derivations of product, quotient, and power rules[edit]Logarithm of a product[edit]Logarithm of a quotient[edit]Logarithm of a power[edit]Changing the base[edit]Proof\/derivation[edit]Summation\/subtraction[edit]Exponents[edit]Other\/resulting identities[edit]Inequalities[edit]Calculus identities[edit]Limits[edit]Derivatives of logarithmic functions[edit]Integral definition[edit]Integrals of logarithmic functions[edit]Approximating large numbers[edit]Complex logarithm identities[edit]Definitions[edit]Constants[edit]Summation[edit]Powers[edit]See also[edit]References[edit]External links[edit]Trivial identities[edit]       (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4logb\u2061(1)=0{displaystyle log _{b}(1)=0}becauseb0=1{displaystyle b^{0}=1}logb\u2061(b)=1{displaystyle log _{b}(b)=1}becauseb1=b{displaystyle b^{1}=b}Explanations[edit]By definition, we know that:       (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4logb(y)=x\u27fabx=y{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}},where b\u22600{displaystyle color {blue}bcolor {black}neq 0} or b\u22601{displaystyle color {blue}bcolor {black}neq 1}.Setting x=0{displaystyle color {red}xcolor {black}=0},we can see that:bx=y\u27fab(0)=y\u27fa1=y\u27fay=1{displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{(0)}color {black}=color {green}ycolor {black}iff color {blue}1color {black}=color {green}ycolor {black}iff color {green}ycolor {black}=color {blue}1color {black}}. So, substituting these values into the formula, we see that:logb(y)=x\u27falogb(1)=0{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}(color {blue}1color {black})=color {red}0color {black}}, which gets us the first property.Setting x=1{displaystyle color {red}xcolor {black}=1},we can see that:bx=y\u27fab(1)=y\u27fab=y\u27fay=b{displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{(1)}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}=color {green}ycolor {black}iff color {green}ycolor {black}=color {blue}bcolor {black}}. So, substituting these values into the formula, we see that:logb(y)=x\u27falogb(b)=1{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}(color {blue}bcolor {black})=color {red}1color {black}}, which gets us the second property.Many mathematical identities are called  trivial , only because they are relatively simple (typically from the perspective of an experienced mathematician).This is not to say that calling an identity or formula  trivial  means that it’s not important.Cancelling exponentials[edit]Logarithms and exponentials with the same base cancel each other. This is true because logarithms and exponentials are inverse operations\u2014much like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations.blogb\u2061(x)=x\u00a0because\u00a0antilogb(logb\u2061(x))=x{displaystyle b^{log _{b}(x)}=x{text{ because }}{mbox{antilog}}_{b}(log _{b}(x))=x}logb\u2061(bx)=x\u00a0because\u00a0logb\u2061(antilogb(x))=x{displaystyle log _{b}(b^{x})=x{text{ because }}log _{b}({mbox{antilog}}_{b}(x))=x}[1]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 color {red}xcolor {black}} and x{displaystyle x} may not be referring to the same number)logb(y)=x\u27fabx=y{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}Looking at the equation bx=y{displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}, and substituting the value for x{displaystyle color {red}xcolor {black}} oflogb(y)=x{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}}, we get the following equation:bx=y\u27fablogb\u2061(y)=y\u27fablogb(y)=y{displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{log _{b}(y)}color {black}=color {green}ycolor {black}iff color {blue}bcolor {black}color {red}^{color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})}color {black}=color {green}ycolor {black}}, which gets us the first equation.Another more rough way to think about it is that bsomething=y{displaystyle color {blue}bcolor {black}color {red}^{text{something}}color {black}=color {green}ycolor {black}},and that that “something{displaystyle color {red}{text{something}}}” is logb(y){displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})}.Looking at the equation logb(y)=x{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}}, and substituting the value for y{displaystyle color {green}ycolor {black}} of bx=y{displaystyle color {blue}bcolor {black}color {red}^{x}color {black}=color {green}ycolor {black}}, we get the following equation:logb(y)=x\u27falogb(bx)=x\u27falogb(bx)=x{displaystyle color {black}log color {blue}_{b}color {black}(color {green}ycolor {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}(color {green}b^{x}color {black})=color {red}xcolor {black}iff color {black}log color {blue}_{b}color {black}({color {blue}bcolor {black}color {red}^{x}color {black}}color {black})=color {red}xcolor {black}}, which gets us the second equation.Another more rough way to think about it is that logb(something)=x{displaystyle color {black}log color {blue}_{b}color {black}(color {green}{text{something}}color {black})=color {red}xcolor {black}},and that that something “something{displaystyle color {green}{text{something}}}” is bx{displaystyle color {blue}bcolor {black}color {red}^{x}color {black}}.Using simpler operations[edit]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.[2] 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\u22601{displaystyle bneq 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:xy=blogb\u2061(x)blogb\u2061(y)=blogb\u2061(x)+logb\u2061(y)\u21d2logb\u2061(xy)=logb\u2061(blogb\u2061(x)+logb\u2061(y))=logb\u2061(x)+logb\u2061(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:xy=(blogb\u2061(x))y=bylogb\u2061(x)\u21d2logb\u2061(xy)=ylogb\u2061(x){displaystyle x^{y}=(b^{log _{b}(x)})^{y}=b^{ylog _{b}(x)}Rightarrow log _{b}(x^{y})=ylog _{b}(x)}The law relating to quotients then follows:logb\u2061(xy)=logb\u2061(xy\u22121)=logb\u2061(x)+logb\u2061(y\u22121)=logb\u2061(x)\u2212logb\u2061(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)}logb\u2061(1y)=logb\u2061(y\u22121)=\u2212logb\u2061(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:logb\u2061(xy)=logb\u2061(x1y)=1ylogb\u2061(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[edit]These are the three main logarithm laws\/rules\/principles,[3] from which the other properties listed above can be proven. Each of these logarithm properties correspond to their respective exponent law, and their derivations\/proofs will hinge on those facts. There are multiple ways to derive\/prove each logarithm law \u2013 this is just one possible method.Logarithm of a product[edit]To state the logarithm of a product law formally:\u2200b\u2208R+,b\u22601,\u2200x,y,\u2208R+,logb\u2061(xy)=logb\u2061(x)+logb\u2061(y){displaystyle forall bin mathbb {R} _{+},bneq 1,forall x,y,in mathbb {R} _{+},log _{b}(xy)=log _{b}(x)+log _{b}(y)}Derivation:Let b\u2208R+{displaystyle bin mathbb {R} _{+}}, where b\u22601{displaystyle bneq 1},and let x,y\u2208R+{displaystyle x,yin mathbb {R} _{+}}. We want to relate the expressions logb\u2061(x){displaystyle log _{b}(x)} and logb\u2061(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 logb\u2061(x){displaystyle log _{b}(x)} and logb\u2061(y){displaystyle log _{b}(y)} quite often, we will give them some variable names to make working with them easier: Let m=logb\u2061(x){displaystyle m=log _{b}(x)}, and let n=logb\u2061(y){displaystyle n=log _{b}(y)}.Rewriting these as exponentials, we see that m=logb\u2061(x)\u27fabm=x{displaystyle m=log _{b}(x)iff b^{m}=x} and n=logb\u2061(y)\u27fabn=y{displaystyle n=log _{b}(y)iff b^{n}=y}. From here, we can relate bm{displaystyle b^{m}} (i.e. x{displaystyle x}) and bn{displaystyle b^{n}} (i.e. y{displaystyle y}) using exponent laws asxy=(bm)(bn)=bm\u22c5bn=bm+n{displaystyle xy=(b^{m})(b^{n})=b^{m}cdot b^{n}=b^{m+n}}To recover the logarithms, we apply logb{displaystyle log _{b}} to both sides of the equality.logb\u2061(xy)=logb\u2061(bm+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 logb\u2061(bm+n)=m+n{displaystyle log _{b}(b^{m+n})=m+n}, givinglogb\u2061(xy)=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}.logb\u2061(xy)=logb\u2061(x)+logb\u2061(y){displaystyle log _{b}(xy)=log _{b}(x)+log _{b}(y)}This completes the derivation.Logarithm of a quotient[edit]To state the logarithm of a quotient law formally:\u2200b\u2208R+,b\u22601,\u2200x,y,\u2208R+,logb\u2061(xy)=logb\u2061(x)\u2212logb\u2061(y){displaystyle forall bin mathbb {R} _{+},bneq 1,forall x,y,in mathbb {R} _{+},log _{b}left({frac {x}{y}}right)=log _{b}(x)-log _{b}(y)}Derivation:Let b\u2208R+{displaystyle bin mathbb {R} _{+}}, where b\u22601{displaystyle bneq 1},and let x,y\u2208R+{displaystyle x,yin mathbb {R} _{+}}.We want to relate the expressions logb\u2061(x){displaystyle log _{b}(x)} and logb\u2061(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 logb\u2061(x){displaystyle log _{b}(x)} and logb\u2061(y){displaystyle log _{b}(y)} quite often, we will give them some variable names to make working with them easier: Let m=logb\u2061(x){displaystyle m=log _{b}(x)}, and let n=logb\u2061(y){displaystyle n=log _{b}(y)}.Rewriting these as exponentials, we see that: m=logb\u2061(x)\u27fabm=x{displaystyle m=log _{b}(x)iff b^{m}=x} and n=logb\u2061(y)\u27fabn=y{displaystyle n=log _{b}(y)iff b^{n}=y}. From here, we can relate bm{displaystyle b^{m}} (i.e. x{displaystyle x}) and bn{displaystyle b^{n}} (i.e. y{displaystyle y}) using exponent laws asxy=(bm)(bn)=bmbn=bm\u2212n{displaystyle {frac {x}{y}}={frac {(b^{m})}{(b^{n})}}={frac {b^{m}}{b^{n}}}=b^{m-n}}To recover the logarithms, we apply logb{displaystyle log _{b}} to both sides of the equality.logb\u2061(xy)=logb\u2061(bm\u2212n){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 logb\u2061(bm\u2212n)=m\u2212n{displaystyle log _{b}(b^{m-n})=m-n}, givinglogb\u2061(xy)=m\u2212n{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}.logb\u2061(xy)=logb\u2061(x)\u2212logb\u2061(y){displaystyle log _{b}left({frac {x}{y}}right)=log _{b}(x)-log _{b}(y)}This completes the derivation.Logarithm of a power[edit]To state the logarithm of a power law formally,\u2200b\u2208R+,b\u22601,\u2200x\u2208R+,\u2200r\u2208R,logb\u2061(xr)=rlogb\u2061(x){displaystyle forall bin mathbb {R} _{+},bneq 1,forall xin mathbb {R} _{+},forall rin mathbb {R} ,log _{b}(x^{r})=rlog _{b}(x)}Derivation:Let b\u2208R+{displaystyle bin mathbb {R} _{+}}, where b\u22601{displaystyle bneq 1}, let x\u2208R+{displaystyle xin mathbb {R} _{+}}, and let r\u2208R{displaystyle rin mathbb {R} }. For this derivation, we want to simplify the expression logb\u2061(xr){displaystyle log _{b}(x^{r})}. To do this, we begin with the simpler expression logb\u2061(x){displaystyle log _{b}(x)}. Since we will be using logb\u2061(x){displaystyle log _{b}(x)} often, we will define it as a new variable: Let m=logb\u2061(x){displaystyle m=log _{b}(x)}.To more easily manipulate the expression, we rewrite it as an exponential. By definition, m=logb\u2061(x)\u27fabm=x{displaystyle m=log _{b}(x)iff b^{m}=x}, so we havebm=x{displaystyle b^{m}=x}Similar to the derivations above, we take advantage of another exponent law. In order to have xr{displaystyle x^{r}} in our final expression, we raise both sides of the equality to the power of r{displaystyle r}:(bm)r=(x)rbmr=xr{displaystyle {begin{aligned}(b^{m})^{r}&=(x)^{r}\\b^{mr}&=x^{r}end{aligned}}}where we used the exponent law (bm)r=bmr{displaystyle (b^{m})^{r}=b^{mr}}.To recover the logarithms, we apply logb{displaystyle log _{b}} to both sides of the equality.logb\u2061(bmr)=logb\u2061(xr){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 logb\u2061(bmr)=mr{displaystyle log _{b}(b^{mr})=mr}.mr=logb\u2061(xr){displaystyle mr=log _{b}(x^{r})}Substituting in the original value for m{displaystyle m}, rearranging, and simplifying gives(logb\u2061(x))r=logb\u2061(xr)rlogb\u2061(x)=logb\u2061(xr)logb\u2061(xr)=rlogb\u2061(x){displaystyle {begin{aligned}left(log _{b}(x)right)r&=log _{b}(x^{r})\\rlog _{b}(x)&=log _{b}(x^{r})\\log _{b}(x^{r})&=rlog _{b}(x)end{aligned}}}This completes the derivation.Changing the base[edit]To state the change of base logarithm formula formally:\u2200a,b\u2208R+,a,b\u22601\u2200x\u2208R+,logb\u2061(x)=loga\u2061(x)loga\u2061(b){displaystyle forall a,bin mathbb {R} _{+},a,bneq 1forall xin mathbb {R} _{+},log _{b}(x)={frac {log _{a}(x)}{log _{a}(b)}}}This identity is useful to evaluate logarithms on calculators. For instance, most calculators have buttons for ln and for log10, but not all calculators have buttons for the logarithm of an arbitrary base.Proof\/derivation[edit]Let a,b\u2208R+{displaystyle a,bin mathbb {R} _{+}}, where a,b\u22601{displaystyle a,bneq 1} Let x\u2208R+{displaystyle xin mathbb {R} _{+}}. Here, a{displaystyle a} and b{displaystyle b} are the two bases we will be using for the logarithms. They cannot be 1, because the logarithm function is not well defined for the base of 1.[citation needed] The number x{displaystyle x} will be what the logarithm is evaluating, so it must be a positive number. Since we will be dealing with the term logb\u2061(x){displaystyle log _{b}(x)} quite frequently, we define it as a new variable: Let m=logb\u2061(x){displaystyle m=log _{b}(x)}.To more easily manipulate the expression, it can be rewritten as an exponential.bm=x{displaystyle b^{m}=x}Applying loga{displaystyle log _{a}} to both sides of the equality,loga\u2061(bm)=loga\u2061(x){displaystyle log _{a}(b^{m})=log _{a}(x)}Now, using the logarithm of a power property, which states that loga\u2061(bm)=mloga\u2061(b){displaystyle log _{a}(b^{m})=mlog _{a}(b)},mloga\u2061(b)=loga\u2061(x){displaystyle mlog _{a}(b)=log _{a}(x)}Isolating m{displaystyle m}, we get the following:m=loga\u2061(x)loga\u2061(b){displaystyle m={frac {log _{a}(x)}{log _{a}(b)}}}Resubstituting m=logb\u2061(x){displaystyle m=log _{b}(x)} back into the equation,logb\u2061(x)=loga\u2061(x)loga\u2061(b){displaystyle log _{b}(x)={frac {log _{a}(x)}{log _{a}(b)}}}This completes the proof that logb\u2061(x)=loga\u2061(x)loga\u2061(b){displaystyle log _{b}(x)={frac {log _{a}(x)}{log _{a}(b)}}}.This formula has several consequences:logb\u2061a=1loga\u2061b{displaystyle log _{b}a={frac {1}{log _{a}b}}}logbn\u2061a=logb\u2061an{displaystyle log _{b^{n}}a={log _{b}a over n}}bloga\u2061d=dloga\u2061b{displaystyle b^{log _{a}d}=d^{log _{a}b}}\u2212logb\u2061a=logb\u2061(1a)=log1\/b\u2061a{displaystyle -log _{b}a=log _{b}left({1 over a}right)=log _{1\/b}a}logb1\u2061a1\u22eflogbn\u2061an=logb\u03c0(1)\u2061a1\u22eflogb\u03c0(n)\u2061an,{displaystyle log _{b_{1}}a_{1},cdots ,log _{b_{n}}a_{n}=log _{b_{pi (1)}}a_{1},cdots ,log _{b_{pi (n)}}a_{n},}where \u03c0{textstyle pi } is any permutation of the subscripts 1, …, n.  For examplelogb\u2061w\u22c5loga\u2061x\u22c5logd\u2061c\u22c5logd\u2061z=logd\u2061w\u22c5logb\u2061x\u22c5loga\u2061c\u22c5logd\u2061z.{displaystyle log _{b}wcdot log _{a}xcdot log _{d}ccdot log _{d}z=log _{d}wcdot log _{b}xcdot log _{a}ccdot log _{d}z.}Summation\/subtraction[edit]The following summation\/subtraction rule is especially useful in probability theory when one is dealing with a sum of log-probabilities:logb\u2061(a+c)=logb\u2061a+logb\u2061(1+ca){displaystyle log _{b}(a+c)=log _{b}a+log _{b}left(1+{frac {c}{a}}right)}because(a+c)=a\u00d7(1+ca){displaystyle left(a+cright)=atimes left(1+{frac {c}{a}}right)}logb\u2061(a\u2212c)=logb\u2061a+logb\u2061(1\u2212ca){displaystyle log _{b}(a-c)=log _{b}a+log _{b}left(1-{frac {c}{a}}right)}because(a\u2212c)=a\u00d7(1\u2212ca){displaystyle left(a-cright)=atimes left(1-{frac {c}{a}}right)}Note that the subtraction identity is not defined if a=c{displaystyle a=c}, since the logarithm of zero is not defined. Also note that, when programming, a{displaystyle a} and c{displaystyle c} may have to be switched on the right hand side of the equations if c\u226ba{displaystyle cgg a} to avoid losing the “1 +” due to rounding errors.  Many programming languages have a specific log1p(x) function that calculates loge\u2061(1+x){displaystyle log _{e}(1+x)} without underflow (when x{displaystyle x} is small).More generally:logb\u2061\u2211i=0Nai=logb\u2061a0+logb\u2061(1+\u2211i=1Naia0)=logb\u2061a0+logb\u2061(1+\u2211i=1Nb(logb\u2061ai\u2212logb\u2061a0)){displaystyle log _{b}sum _{i=0}^{N}a_{i}=log _{b}a_{0}+log _{b}left(1+sum _{i=1}^{N}{frac {a_{i}}{a_{0}}}right)=log _{b}a_{0}+log _{b}left(1+sum _{i=1}^{N}b^{left(log _{b}a_{i}-log _{b}a_{0}right)}right)}Exponents[edit]A useful identity involving exponents:xlog\u2061(log\u2061(x))log\u2061(x)=log\u2061(x){displaystyle x^{frac {log(log(x))}{log(x)}}=log(x)}or more universally:xlog\u2061(a)log\u2061(x)=a{displaystyle x^{frac {log(a)}{log(x)}}=a}Other\/resulting identities[edit]11logx\u2061(a)+1logy\u2061(a)=logxy\u2061(a){displaystyle {frac {1}{{frac {1}{log _{x}(a)}}+{frac {1}{log _{y}(a)}}}}=log _{xy}(a)}11logx\u2061(a)\u22121logy\u2061(a)=logxy\u2061(a){displaystyle {frac {1}{{frac {1}{log _{x}(a)}}-{frac {1}{log _{y}(a)}}}}=log _{frac {x}{y}}(a)}Inequalities[edit]Based on,[4][5] and [6]x1+x\u2264ln\u2061(1+x)\u2264x(6+x)6+4x\u2264x\u00a0for all\u00a0\u22121273+2x\u2264x1+x+x2\/12\u2264ln\u2061(1+x)\u2264x1+x\u2264x22+x1+x\u00a0for\u00a00\u2264x, reverse for\u00a0\u221210+loga\u2061(x)=\u2212\u221eif\u00a0a>1{displaystyle lim _{xto 0^{+}}log _{a}(x)=-infty quad {mbox{if }}a>1}"},{"@context":"http:\/\/schema.org\/","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"item":{"@id":"https:\/\/wiki.edu.vn\/en\/wiki14\/#breadcrumbitem","name":"Enzyklop\u00e4die"}},{"@type":"ListItem","position":2,"item":{"@id":"https:\/\/wiki.edu.vn\/en\/wiki14\/list-of-logarithmic-identities-wikipedia\/#breadcrumbitem","name":"List of logarithmic identities – Wikipedia"}}]}]