# Convert Between Logarithmic And Exponential Form

### Learning Objectives

- Convert a logarithm to exponential form
- Convert an exponential expression to logarithmic form

*x*represents the difference in magnitudes on the

**Richter Scale**. How would we solve for

*x*? We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[/latex]. We know that [latex]{10}^{2}=100[/latex] and [latex]{10}^{3}=1000[/latex], so it is clear that

*x*must be some value between 2 and 3, since [latex]y={10}^{x}[/latex] is increasing. We can examine a graph to better estimate the solution. Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above passes the horizontal line test. The exponential function [latex]y={b}^{x}[/latex] is

**one-to-one**, so its inverse, [latex]x={b}^{y}[/latex] is also a function. As is the case with all inverse functions, we simply interchange

*x*and

*y*and solve for

*y*to find the inverse function. To represent

*y*as a function of

*x*, we use a logarithmic function of the form [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex]. The base

*b*

**logarithm**of a number is the exponent by which we must raise

*b*to get that number. We read a logarithmic expression as, "The logarithm with base

*b*of

*x*is equal to

*y*," or, simplified, "log base

*b*of

*x*is

*y*." We can also say, "

*b*raised to the power of

*y*is

*x*," because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[/latex], we can write [latex]{\mathrm{log}}_{2}32=5[/latex]. We read this as "log base 2 of 32 is 5." We can express the relationship between logarithmic form and its corresponding exponential form as follows: [latex-display]{\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1[/latex-display] Note that the base

*b*is always positive. Because logarithm is a function, it is most correctly written as [latex]{\mathrm{log}}_{b}\left(x\right)[/latex], using parentheses to denote function evaluation, just as we would with [latex]f\left(x\right)[/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as [latex]{\mathrm{log}}_{b}x[/latex]. Note that many calculators require parentheses around the

*x*. We can illustrate the notation of logarithms as follows: Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex] and [latex]y={b}^{x}[/latex] are inverse functions.

### A General Note: Definition of the Logarithmic Function

A**logarithm**base

*b*of a positive number

*x*satisfies the following definition. For [latex]x>0,b>0,b\ne 1[/latex], [latex-display]y={\mathrm{log}}_{b}\left(x\right)\text{ is equivalent to }{b}^{y}=x[/latex-display] where,

- we read [latex]{\mathrm{log}}_{b}\left(x\right)[/latex] as, "the logarithm with base
*b*of*x*" or the "log base*b*of*x*." - the logarithm
*y*is the exponent to which*b*must be raised to get*x*.

*x*and

*y*values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,

- the domain of the logarithm function with base [latex]b \text{ is} \left(0,\infty \right)[/latex].
- the range of the logarithm function with base [latex]b \text{ is} \left(-\infty ,\infty \right)[/latex].

### Q & A

#### Can we take the logarithm of a negative number?

*No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.*

### How To: Given an equation in logarithmic form [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex], convert it to exponential form.

- Examine the equation [latex]y={\mathrm{log}}_{b}x[/latex] and identify
*b*,*y*, and*x*. - Rewrite [latex]{\mathrm{log}}_{b}x=y[/latex] as [latex]{b}^{y}=x[/latex].

### Example: Converting from Logarithmic Form to Exponential Form

Write the following logarithmic equations in exponential form.- [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex]
- [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]

Answer:
First, identify the values of *b*, *y*, and *x*. Then, write the equation in the form [latex]{b}^{y}=x[/latex].

- [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] Here, [latex]b=6,y=\frac{1}{2},\text{and } x=\sqrt{6}[/latex]. Therefore, the equation [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] is equivalent to [latex]{6}^{\frac{1}{2}}=\sqrt{6}[/latex].
- [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] Here,
*b*= 3,*y*= 2, and*x*= 9. Therefore, the equation [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] is equivalent to [latex]{3}^{2}=9[/latex].

### Try It

Write the following logarithmic equations in exponential form.- [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]

Answer:

- [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex] is equivalent to [latex]{10}^{6}=1,000,000[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex] is equivalent to [latex]{5}^{2}=25[/latex]

## Convert from exponential to logarithmic form

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base*b*, exponent

*x*, and output

*y*. Then we write [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

### Example: Converting from Exponential Form to Logarithmic Form

Write the following exponential equations in logarithmic form.- [latex]{2}^{3}=8[/latex]
- [latex]{5}^{2}=25[/latex]
- [latex]{10}^{-4}=\frac{1}{10,000}[/latex]

Answer:
First, identify the values of *b*, *y*, and *x*. Then, write the equation in the form [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

- [latex]{2}^{3}=8[/latex] Here,
*b*= 2,*x*= 3, and*y*= 8. Therefore, the equation [latex]{2}^{3}=8[/latex] is equivalent to [latex]{\mathrm{log}}_{2}\left(8\right)=3[/latex]. - [latex]{5}^{2}=25[/latex] Here,
*b*= 5,*x*= 2, and*y*= 25. Therefore, the equation [latex]{5}^{2}=25[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]. - [latex]{10}^{-4}=\frac{1}{10,000}[/latex] Here,
*b*= 10,*x*= –4, and [latex]y=\frac{1}{10,000}[/latex]. Therefore, the equation [latex]{10}^{-4}=\frac{1}{10,000}[/latex] is equivalent to [latex]{\text{log}}_{10}\left(\frac{1}{10,000}\right)=-4[/latex].

### Try It

Write the following exponential equations in logarithmic form.- [latex]{3}^{2}=9[/latex]
- [latex]{5}^{3}=125[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex]

Answer:

- [latex]{3}^{2}=9[/latex] is equivalent to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
- [latex]{5}^{3}=125[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex] is equivalent to [latex]{\text{log}}_{2}\left(\frac{1}{2}\right)=-1[/latex]

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**Authored by:**McClure,Caren.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC-BY + GPL. - College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**Located at:**https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites.**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[email protected].