How To Find The Base of a Logarithm?
Solving for an unknown base of a logarithm simply means working in a reverse form. In this case, you will have a logarithm of a number without knowing the base of that logarithm. From our knowledge of how to evaluate basic logarithm, this reverse process should not pose any challenge to us.
The general notation of a logarithm is
![\[log _{b} y=x\]](data:image/svg+xml;nitro-empty-id=MjY0OjUwMA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNzIgMTYiIHdpZHRoPSI3MiIgaGVpZ2h0PSIxNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
where the small letter b is the base, and the small letter x is the exponent or the power. We can convert this logarithmic equation into an exponential equation to be
![\[y=b^{x}\]](data:image/svg+xml;nitro-empty-id=MjY2OjQ3Mg==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDkgMTgiIHdpZHRoPSI0OSIgaGVpZ2h0PSIxOCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
From here, we can work logarithms either forward or backward, always finding the unknown in the equation.
Example 1
Given that
![\[log _{b} 27=3\]](data:image/svg+xml;nitro-empty-id=MjcwOjUwMw==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgODAgMTciIHdpZHRoPSI4MCIgaGVpZ2h0PSIxNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
find the base of the log.
Solution
We start by rewriting the log into the exponential form.
![\[log _{b} 27=3 \equiv 27=b^{3}\]](data:image/svg+xml;nitro-empty-id=Mjc0OjU5MQ==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTU5IDIxIiB3aWR0aD0iMTU5IiBoZWlnaHQ9IjIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
Next, we solve for b by taking the cube root of both sides of our new equation.
![\[\sqrt[3]{27}=\sqrt[3]{b^{3}}\]](data:image/svg+xml;nitro-empty-id=Mjc2OjU4Nw==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgODcgMTkiIHdpZHRoPSI4NyIgaGVpZ2h0PSIxOSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[3=b\]](data:image/svg+xml;nitro-empty-id=Mjc3OjQ0OA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDAgMTIiIHdpZHRoPSI0MCIgaGVpZ2h0PSIxMiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[log _{3} 27=3\]](data:image/svg+xml;nitro-empty-id=Mjc4OjUwMw==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgODAgMTciIHdpZHRoPSI4MCIgaGVpZ2h0PSIxNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Example 2
Evaluate:
![\[log _{b} 81=-4\]](data:image/svg+xml;nitro-empty-id=MjgxOjUwOA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTQgMTYiIHdpZHRoPSI5NCIgaGVpZ2h0PSIxNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Solution
First write the equation in an exponential form.
![\[81=b^{-4}\]](data:image/svg+xml;nitro-empty-id=Mjg0OjQ4MA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNjggMTciIHdpZHRoPSI2OCIgaGVpZ2h0PSIxNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Second, apply exponential laws:
![\[81=\frac{1}{b^{4}}\]](data:image/svg+xml;nitro-empty-id=Mjg2OjUzMQ==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNTkgMzYiIHdpZHRoPSI1OSIgaGVpZ2h0PSIzNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
We solve for b by taking the 4th root of both sides of the equation.
![\[\sqrt[4]{81}=\sqrt[4]{\frac{1}{b^{4}}}\]](data:image/svg+xml;nitro-empty-id=Mjg4OjY0Mw==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTQgNDMiIHdpZHRoPSI5NCIgaGVpZ2h0PSI0MyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[3=\frac{1}{b}\]](data:image/svg+xml;nitro-empty-id=Mjg5OjUwNA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDQgMzYiIHdpZHRoPSI0NCIgaGVpZ2h0PSIzNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[b=\frac{1}{3}\]](data:image/svg+xml;nitro-empty-id=MjkwOjUwNA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDIgMzYiIHdpZHRoPSI0MiIgaGVpZ2h0PSIzNiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[log _{\frac{1}{3}} 81=-4\]](data:image/svg+xml;nitro-empty-id=MjkxOjU2NA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTggMjIiIHdpZHRoPSI5OCIgaGVpZ2h0PSIyMiIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Example 3
Evaluate:
![\[log _{b} \frac{125}{64}=3\]](data:image/svg+xml;nitro-empty-id=Mjk0OjU2OQ==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTIgMzciIHdpZHRoPSI5MiIgaGVpZ2h0PSIzNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Solution
As usual, we start by rewriting the log equation into an exponential equation.
![\[\frac{125}{64}=b^{3}\]](data:image/svg+xml;nitro-empty-id=Mjk3OjU0MQ==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNjcgMzciIHdpZHRoPSI2NyIgaGVpZ2h0PSIzNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
We solve for b by taking the cube root of both sides of the equation.
![\[\sqrt[3]{\frac{125}{64}}=\sqrt[3]{b^{3}}\]](data:image/svg+xml;nitro-empty-id=Mjk5OjY1NA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTAzIDQzIiB3aWR0aD0iMTAzIiBoZWlnaHQ9IjQzIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg== "Rendered by QuickLaTeX.com")
![\[b=\frac{5}{4}\]](data:image/svg+xml;nitro-empty-id=MzAwOjUwNA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDIgMzciIHdpZHRoPSI0MiIgaGVpZ2h0PSIzNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
![\[log _{\frac{5}{4}} \frac{125}{64}=3\]](data:image/svg+xml;nitro-empty-id=MzAxOjYyNQ==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgOTYgMzciIHdpZHRoPSI5NiIgaGVpZ2h0PSIzNyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4= "Rendered by QuickLaTeX.com")
Remarks
The essential knowledge you need to find the base of a logarithm is the exponential laws and a bit of algebra. From the tutorial, you should be able to evaluate the base of any logarithm from a given equation. In short: Rewrite the logarithmic equation you have into an exponential form and solve for the unknown, which is the base of the logarithm.
About the Author
This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.