Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can I ask how to compute $\log_3 7$, using the changing the base of logarithm.

share|improve this question
3  
$\log_3(7)$ is only an expression. One does not solve expressions. One solves equations like $\log_3(7) = x - 2$ (for example). Is something missing? –  JavaMan May 2 '11 at 15:10
    
If you mean evaluate in terms of logarithms that appear on your calculator, you want to use the change of base formula to write it in terms of $\ln$ or $\log_{10}$. Do you know what the change of base formula is? –  Jonas Meyer May 2 '11 at 15:14
1  
@DJC: My guess is that the question is meant to be "how can I compute $\log_3(7)$" using the change of base of logarithm (e.g., if you only know how to compute natural or common logarithms...) –  Arturo Magidin May 2 '11 at 15:14
    
sorry for my bad english, i should use find the values of instead –  dramasea May 2 '11 at 15:15
2  
This is a duplicate of at least a few other old questions. –  Bill Dubuque May 2 '11 at 16:20

1 Answer 1

up vote 11 down vote accepted

If you mean, "How can I calculate $\log_3 7$ using the change of base formula?":

I've never memorized the change of base formula, I always re-derive it as needed. The key is to remember what the expression means: $\log_3 7 = r$ means that $3^r = 7$. Taking logarithms base $b$ on both sides, we have $$\begin{align*} 3^r &= 7\\ \log_b(3^r) &= \log_b(7)\\ r\log_b 3&= \log_b 7\\ r &= \frac{\log_b 7}{\log_b 3}\\ \log_3 7 &= \frac{\log_b 7}{\log_b 3}. \end{align*}$$ So if you want to compute $\log_3 7$ using the natural log, you would have $$\log_3 7 = \frac{\ln 7}{\ln 3}.$$ If you want to compute them using the common logarithm (base 10), you would compute $$\log_3 7 = \frac{\log 7}{\log 3}.$$

share|improve this answer
    
+1 for re-deriving basic log properties. –  Hans Parshall May 2 '11 at 15:28
    
@Hans: Heh; I honestly don't consider the change-of-base formula as a "basic property". To me, the "basic properties" are $\log_a(b) = r \Leftrightarrow a^r=b$ (definition), and the corresponding equalities that are derived from the basic properties of the exponential: $a^ra^s = a^{r+s}$ and $(a^r)^s = a^{rs}$ –  Arturo Magidin May 2 '11 at 16:07
    
All the same, from experience, I've found that each of these things are worth deriving in front of students needing help. Logs seem to be especially mysterious to even competent calculus students. The derivations can help demystify. –  Hans Parshall May 2 '11 at 16:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.