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Change-of-base Formula Made Easy

It is just another day in solving log equations, but you are challenged with a difficult one, which includes logs with different bases. How should you proceed to solve it? It is easy; you just need to use the change-of-base formula. This special formula helps you rewrite your different logs in the same base, enabling you to find solutions for complicated log equations. It also unveils that logs in different bases are proportional to each other, this means that all logs are the same function, they just differ in scale, but their functional behaviors are identical, and they all obey the constraint log⁡(1)=0\log (1) = 0.

Would you like to know more about logs? Then access our articles titled "Log Rules Made Simple: Understanding the Laws of Logarithms" and "How to Solve Log Equations Like a Pro". We are sure they will help you to become a log expert.

Along with this article, we are going to show you:

  • The change-of-base formula.
  • How to apply the change-of-base formula for logs.
  • How to derive the logarithm change-of-base formula.
  • Change-of-base formula applied to log equations.
  • How to use the change-of-base formula with a calculator.
  • Why should I use the change-of-base formula?
  • How can I change logs between bases?.
  • And much more.

Come with us to master the log change-of-base formula and see the logarithms from a new perspective.

As the name suggests, the change-of-base formula is used to change the base of a given logarithm. It is a handy formula that you can apply to write log⁡10\log_{10} in terms of ln⁡\ln, and also to simplify log equations. The change-of-basis formula is given by:

log⁡a(x)=log⁡b(x)log⁡b(a)\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}

Therefore, the logarithm change of base formula unveils that logs with different bases are proportional to each other, or in other words, we have:

log⁡a(x)=k log⁡b(x)k=1log⁡b(a)\small \log_{a}(x) = k\,\log_{b}(x) \qquad k = \frac{1}{\log_{b}(a)}

The previous equation means that logs with different bases are essentially the same function, just shifted by a scale. Now, you can apply such a formula to change between the natural logarithm (ln⁡\ln) and the common logarithm (log⁡10\log_{10}). These scalings follow below:

ln⁡(x)=log⁡10(x)log⁡10(e)log⁡10(x)=ln⁡(x)ln⁡(10)\small \ln (x) = \frac{\log_{10}(x)}{\log_{10}(e)} \qquad \log_{10} (x) = \frac{\ln(x)}{\ln(10)}

Now that you know the log change-of-base formula, we can apply it with some examples.

  1. What is the value of log⁡64(8)\log_{64}(8)?

    We can answer this question easily by applying the change-of-base formula, as follows:

log⁡64(8)=log⁡10(8)log⁡10(64)=log⁡10(8)log⁡10(82)=log⁡10(8)2⋅log⁡10(8)=12\quad\begin{split} \log_{64}(8) &= \frac{\log_{10}(8)}{\log_{10}(64)} \\[1em] & = \frac{\log_{10}(8)}{\log_{10}(8^2)} \\[1em] & = \frac{\log_{10}(8)}{2\cdot \log_{10}(8)} \\[1em] & = \frac{1}{2} \end{split}

  1. Evaluate log⁡2(50)\log_{2}(50).

    By using the change-of-base formula, we can rewrite this log in terms of the natural logarithm:

log⁡2(50)=ln⁡(50)ln⁡(2)\quad\log_{2}(50) = \frac{\ln(50)}{\ln(2)}

🙋 You can compute these logarithms with our amazing natural log calculator 🇺🇸, yielding to:

log⁡2(50)=3.9120.693=5.645\quad\begin{split} \log_{2}(50) &= \frac{3.912}{0.693} \\[1em] & = 5.645 \end{split}

The logarithm change-of-base formula is derived using the connection between log⁡\log and the exponential. You should remember that log⁡\log and exponential are inverse functions, which means that:

log⁡a(x)=yay=x\log_a (x) = y \qquad a^y = x

Thus, let's rewrite the equation of the right-hand side by taking the log⁡b\log_{b} of both sides and simplifying, resulting in:

log⁡b(ay)=log⁡b(x)y⋅log⁡b(a)=log⁡b(x)\begin{split} & \log_{b} (a^y) = \log_{b}(x) \\[1em] & y\cdot \log_{b}(a) = \log_{b}(x) \\[1em] \end{split}

then, by substituting y=log⁡a(x)y = \log_{a}(x) in the previous expression, and after some simplifications, we finally obtain the change-of-base formula:

log⁡a(x)⋅log⁡b(a)=log⁡b(x)log⁡a(x)=log⁡b(x)log⁡b(a)\begin{split} & \log_{a}(x) \cdot \log_{b}(a) = \log_{b}(x)\\[1em] & \log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)} \end{split}

The change-of-base formula is also applied to simplify log equations. This process allows us to find solutions for complex log equations, and it is a powerful application of the change-of-base formula. Let's see with some examples:

1. What is the value of xx if log⁡3(2x)=log⁡9(x)+1\log_{3}(2x) = \log_{9}(x)+1?

In order to find the solution, first, we use the log change-of-base formula to rewrite log⁡9(x)\log_{9}(x):

log⁡9(x)=log⁡3(x)log⁡3(9)=log⁡3(x)log⁡3(32)=log⁡3(x)2⋅log⁡3(3)=log⁡3(x)2\begin{split} \log_{9}(x) &= \frac{\log_{3}(x)}{\log_{3}(9)} \\[1em] & =\frac{\log_{3}(x)}{\log_{3}(3^2)} \\[1em] & = \frac{\log_{3}(x)}{2\cdot \log_{3}(3)} \\[1em] & = \frac{\log_{3}(x)}{2} \end{split}

Now, we can substitute the last expression into our log equation, and determine the value of xx:

log⁡3(2x)=log⁡3(x)2+1log⁡3(2)+log⁡3(x)=log⁡3(x)2+1log⁡3(x)2=1−log⁡3(2)log⁡3(x)=2(1−log⁡3(2))x=32(1−log⁡3(2))x=9⋅3log⁡3(2)−2x=9⋅2−2x=94\begin{split} & \log_{3}(2x) = \frac{\log_{3}(x)}{2}+1 \\[1em] & \log_{3}(2) + \log_{3}(x) =\frac{\log_{3}(x)}{2}+1 \\[1em] & \frac{\log_{3}(x)}{2} = 1 - \log_{3}(2) \\[1em] & \log_{3}(x) =2(1-\log_{3}(2)) \\[1em] & x = 3^{2(1-\log_{3}(2))} \\[1em] & x = 9\cdot 3^{\log_{3}(2)^{-2}} \\[1em] & x = 9 \cdot 2^{-2} \\[1em] & x = \frac{9}{4} \end{split}

2. Evaluate xx in log⁡5(x−1)=log⁡25(4x)\log_{5}(x-1) = \log_{25}(4x).

We start to solve this problem by applying the logarithm change-of-base formula to log⁡25(4x)\log_{25}(4x):

log⁡25(4x)=log⁡5(x)log⁡5(25)=log⁡5(x)2\begin{split} \log_{25}(4x) &= \frac{\log_{5}(x)}{\log_{5}(25)} \\[1em] & =\frac{\log_{5}(x)}{2} \end{split}

By substituting the previous result in the log equation, we have

log⁡5(x−1)=log⁡5(x)2log⁡5(x−1)2=log⁡5(x)x2−6x+1=0x=3+22\begin{split} & \log_{5}(x-1) = \frac{\log_{5}(x)}{2}\\[1em] & \log_{5}(x-1)^2 = \log_{5}(x)\\[1em] & x^2-6x+1 = 0 \\[1em] & x = 3 + 2\sqrt{2} \end{split}

We know that it is not an easy task to calculate all these logs by hand. Then, we are here to help you by bringing interesting tools to simplify the computation of your logarithms. Take a look at our dedicated tools:

Feel free to check them and to apply the change-of-base formula for logs easily.

The change-of-base formula is an equation used to change the base of a logarithm. It is a method used to simplify log equations and also reveals that logarithms differ only by a scale constant, meaning that logs have the same function no matter what base you choose.

You should use the logarithm change-of-base formula to simplify your log equations with mixed bases, simplify products of logs, and change between the natural log and the common log.

You apply the change-of-base formula to change logs between bases and determine the value of a complex log. For instance, suppose that you want to compute log5(12). In order to calculate it, follow the steps below:

  1. Choose a proper base to make your calculation: in this case, we will work with the natural log.

  2. Apply the change-of-base formula:
    log5(12) = ln(12)/ln(5)

  3. Use a log calculator to determine the numerical values of the logs:

    ln(12) = 2.485 and ln(5) = 1.609

  4. Calculate the fraction and derive the value of your log:

    log5(12) = 1.544

This article was written by João Rafael Lucio dos Santos and reviewed by Steven Wooding.