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Exponentials and logarithms

Laws of logarithms

Laws of logarithms

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Summary

Laws of logarithms

In a nutshell

Similar to the laws for exponentials, e.g. ax×ay=ax+ya^x \times a^y = a^{x + y}ax×ay=ax+y, laws for logarithms exist which you can use to manipulate expressions they feature in. Here you will derive and cover those laws.



Multiplication, division and power laws

Multiplication law

Let aaa be a positive constant not equal to 111​, and let mmm​ and nnn​ be any real values. Suppose that am=xa^m = xam=x​, and an=ya^n = yan=y​. Then the laws of exponentials tell you that:


​xy=am×an=am+n\begin{aligned}xy &= a^m \times a^n\\&= a^{m+n}\end{aligned}xy​=am×an=am+n​​


Applying log⁡a()\log_a()loga​()​ to xxx​, yyy, and xyxyxy​, you have that:


​m=log⁡a(x)n=log⁡a(y)m+n=log⁡a(xy)\begin{aligned}m &= \log_a(x)\\n &= \log_a(y)\\m+n &= \log_a(xy)\end{aligned}mnm+n​=loga​(x)=loga​(y)=loga​(xy)​​​


Put these together to derive the multiplication law for logarithms:


​log⁡a(xy)=log⁡a(x)+log⁡a(y)\boxed{\log_a(xy) = \log_a(x) + \log_a(y)}loga​(xy)=loga​(x)+loga​(y)​​​


Division law

If xxx and yyy are constants with y≠0y \neq 0y=0, then you have that xy×y=x\dfrac{x}{y} \times y = xyx​×y=x. Apply log⁡a()\log_a()loga​() to this equation, use the multiplication law:


​log⁡a(xy×y)=log⁡a(x)log⁡a(xy)+log⁡a(y)=log⁡a(x)\begin{aligned}\log_a\left(\dfrac{x}{y} \times y \right) &= \log_a(x)\\\log_a\left(\dfrac{x}{y}\right) + \log_a(y) &= \log_a(x)\end{aligned} loga​(yx​×y)loga​(yx​)+loga​(y)​=loga​(x)=loga​(x)​​​


Rearrange this to derive the division law for logarithms:


​log⁡a(xy)=log⁡a(x)−log⁡a(y)\boxed{\log_a\left(\dfrac{x}{y}\right)= \log_a(x) - \log_a(y)} loga​(yx​)=loga​(x)−loga​(y)​​​


Power law

You have that x=amx = a^mx=am. Laws of exponentials tell you that:


​xk=(am)k=akm\begin{aligned}x^k &= (a^m)^k\\&= a^{km}\end{aligned}xk​=(am)k=akm​​​


Apply log⁡a()\log_a()loga​()​ to get:


​log⁡a(xk)=km\log_a(x^k) = kmloga​(xk)=km​​


Substitute m=log⁡a(x)m = \log_a(x)m=loga​(x)​ to derive the power law for logarithms:


​log⁡a(xk)=klog⁡a(x)\boxed{\log_a(x^k) = k\log_a(x)}loga​(xk)=kloga​(x)​​​



Example 1

Let aaa​ be a positive constant not equal to 111​, and let x≥0x \ge 0x≥0​. Show that log⁡a(x)=log⁡a(x)2\log_a(\sqrt{x}) = \dfrac{\log_a(x)}{2}loga​(x​)=2loga​(x)​.


You have that x=x12\sqrt{x} = x^{\frac12}x​=x21​. The power law for logarithms tells you that:


​x=x12log⁡a(x)=log⁡a(x12)log⁡a(x)=12log⁡a(x)\begin{aligned}\sqrt{x} &= x^{\frac12}\\\log_a(\sqrt{x}) &= \log_a(x^\frac12)\\\log_a(\sqrt{x}) &= \dfrac12\log_a(x)\end{aligned}x​loga​(x​)loga​(x​)​=x21​=loga​(x21​)=21​loga​(x)​​​


Therefore, log⁡a(x)=log⁡a(x)2‾\underline{\log_a(\sqrt{x}) = \dfrac{\log_a(x)}{2}}loga​(x​)=2loga​(x)​​.


​

Special cases

For a positive constant aaa​ not equal to 111​, recall that the unique value of xxx​ such that ax=1a^x = 1ax=1​ is x=0x = 0x=0. Similarly, the unique value of xxx​ for which ax=aa^x = aax=a​ is x=1x = 1x=1​. Therefore, you have the following special cases:


​log⁡a(1)=0log⁡a(a)=1\begin{aligned}\log_a(1) &= 0\\\log_a(a) &= 1\end{aligned}loga​(1)loga​(a)​=0=1​​​

 Using this and the division law, you can obtain the following useful case:


​log⁡a(1x)=log⁡a(1)−log⁡a(x)=−log⁡a(x)\begin{aligned}\log_a\left(\dfrac1x\right) & = \log_a(1) - \log_a(x)\\&= -\log_a(x)\end{aligned}loga​(x1​)​=loga​(1)−loga​(x)=−loga​(x)​​​



Example 2

WIthout using a calculator, find the exact value of log⁡3(279)\log_3\left(\dfrac{\sqrt{27}}{9}\right)log3​(927​​).


You have that 27=332\sqrt{27} = 3^{\frac32}27​=323​, and 9=329 = 3^29=32. Use the power and division laws to get:


log⁡3(279)=log⁡3(27)−log⁡3(9)=log⁡3(332)−log⁡3(32)=32log⁡3(3)−2log⁡3(3)=32−2=−12\begin{aligned}\log_3\left(\dfrac{\sqrt{27}}{9}\right)&= \log_3(\sqrt{27}) - \log_3(9)\\&= \log_3(3^{\frac32}) - \log_3(3^2)\\&= \dfrac32\log_3(3) - 2\log_3(3)\\&= \dfrac{3}{2} - 2\\&= -\dfrac12\end{aligned}log3​(927​​)​=log3​(27​)−log3​(9)=log3​(323​)−log3​(32)=23​log3​(3)−2log3​(3)=23​−2=−21​​​​


Therefore, the exact value of log⁡3(279)\log_3\left(\dfrac{\sqrt{27}}{9}\right)log3​(927​​) is −12‾\underline{-\dfrac12}−21​​.


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FAQs - Frequently Asked Questions

What is the power law for logarithms?

log(x^k) = klog(x)

What is the division law for logarithms?

log(x/y) = log(x) - log(y)

What is the multiplication law for logarithms?

log(xy) = log(x) + log(y)

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