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Natural logarithms

Natural logarithms

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Summary

Natural logarithms

​​In a nutshell

The natural logarithm, ln⁡(x)\ln(x)​, is the logarithm with base ee​, and therefore the functions exe^x​ and ln⁡(x)\ln(x)​ are inverses to each other.



The graph of the equation y=ln⁡(x)y = \ln(x)​​

A consequence of the functions exe^x​ and ln⁡(x)\ln(x)​ being inverse to each other is that the graphs of the equations y=exy = e^x​ and y=ln⁡(x)y = \ln(x)​ are reflections of each other about the line y=xy = x. This is because eln⁡(x)=ln⁡(ex)=xe^{\ln(x)} = \ln(e^{x}) = x​, and so if (a,b)=(a,ea)(a,b) = (a,e^a)​ lies on the graph of the equation y=exy = e^x​, then the point (b,ln⁡(b))=(b,ln⁡(ea))=(b,a)(b, \ln(b)) = (b, \ln(e^a)) = (b,a)​ lies on the graph of the equation y=ln⁡(x)y = \ln(x)​.

Maths; Exponentials and logarithms; KS5 Year 12; Natural logarithms

As xx​ decreases, exe^x​ tends towards 00​. Consequently, the graph of the equation y=ln⁡(x)y = \ln(x)​ has the yy​-axis as an asymptote. This means that ln⁡(x)\ln(x)​ is only defined when xx is a positive real number.


​e0=1e^0 = 1​, and so ln⁡(1)=0\ln(1) = 0​. The graph of the equation y=ln⁡(x)y = \ln(x)​ therefore passes through the point (1,0)(1,0)​.


Example 1

Solve the equation ln⁡(x−5)+ln⁡(x+2)=ln⁡(18)\ln(x-5) + \ln(x+2) = \ln(18)​


Use the multiplication law for logarithms to get that ln⁡(x−5)+ln⁡(x+2)=ln⁡((x−5)(x+2))\ln(x-5) + \ln(x+2) = \ln((x-5)(x+2)). Expand this, and raise ee​ to the power of both sides of the equation:


ln⁡((x−5)(x+2))=ln⁡(18)ln⁡(x2−3x−10)=ln⁡(18)eln⁡(x2−3x−10)=eln⁡(18)x2−3x−10=18x2−3x−28=0(x−7)(x+4)=0\begin{aligned}\ln((x-5)(x+2)) &= \ln(18)\\\ln(x^2 - 3x - 10) &= \ln(18)\\e^{\ln(x^2 - 3x - 10)} &= e^{\ln(18)}\\x^2 - 3x - 10 &= 18\\x^2 - 3x - 28 &= 0\\(x-7)(x+4) &= 0\end{aligned}​​

It looks like the solutions to the equation are x=7x = 7 and x=−4x = -4. However, ln⁡(x−5)\ln(x-5) is only defined when x>5x \gt 5, and ln⁡(x+2)\ln(x+2)​ is only defined when x>−2x \gt -2​. So, ln⁡(x−5)+ln⁡(x+2)\ln(x-5) + \ln(x+2)​ is only defined when x>5x \gt 5​.


Therefore, the only solution to the equation ln⁡(x−5)+ln⁡(x+2)=ln⁡(18)\ln(x-5) + \ln(x+2) = \ln(18) is x=7‾\underline{x=7}.


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Exponential functions

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Logarithms

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Natural logarithms

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