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Suppose I want to solve $3^x=10$. I convert it to logarithmic form $$\log_{3}10=x$$ then change bases $$\frac{\log10}{\log3}=x$$ and this will yield the same answer as if I had written $$\frac{\ln10}{\ln3}=x$$ But log and ln have different bases. Why does this work both ways?

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  • $\begingroup$ Your first "inserted" equation is wrong. It should be $x = \log_3 10$, not $x=\log_{10} 3$. Generally, $a^b=c \iff b=\log_a c$. $\endgroup$
    – MPW
    Sep 1 '16 at 16:54
  • $\begingroup$ @MPW Right, I had to edit the question. And the second half was just rambling about the first half. $\endgroup$ Sep 1 '16 at 16:59
  • $\begingroup$ @ParclyTaxel Thank you very much for the concise summary. $\endgroup$
    – Raboush2
    Sep 1 '16 at 17:08
  • $\begingroup$ Mathematicians writing "$\log$" usually mean the same thing as "$\ln$". The convention used by many engineers, biologists, astronomers, and others, that $\log$ means the base-$10$ logarithm, was to a large extent rendered obsolete by in advent of calculators around 1970 or so, but it is perpetuated by calculators. Base-10 logarithms are useful in computations by hand because you only need a table going from $1$ to $10$: If you want the logarithm of $145$, you find the logarithm of $1.45$ in the table and then add $2$ to it since the decimal point is $2$ places to the right of there. $\qquad$ $\endgroup$ Sep 1 '16 at 17:20
  • $\begingroup$ It used to be that if you wanted to tell students how to find things like $\log_3 20$ with a calculator, you told them it's $\dfrac{\log_{10} 20}{\log_{10} 3}$ or else it's $\dfrac{\ln 20}{\ln 3}$, and they got base-$10$ or base-$e$ logarithms from the calculator. But today you find that some of them have calculators on which you enter both the base and the argument. (I haven't yet encountered on that gives GCDs. Maybe that means I'm not up to date?) $\qquad$ $\endgroup$ Sep 1 '16 at 17:22
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You can use whatever basis: the equality $3^x=10$ is equivalent to $$ \log_a 3^x=\log_a 10 $$ for any $a>0$, $a\ne1$. Since $\log_a 3^x=x\log_a 3$, we get $$ x=\frac{\log_a 10}{\log_a 3} $$ so you see that the final result is independent of the base.

You can get the change of base formula by considering $b^c=k$ that says $c=\log_b k$; but we also have $c\log_a b=\log_a k$ and therefore $$ \log_b k=\frac{\log_a k}{\log_a b} $$ In the case of $b=3$, $k=10$ and $a=e$, you get $$ \log_3 10=\frac{\ln 10}{\ln 3} $$

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    $\begingroup$ Perhaps more to the point is the fact that $$\frac{\log_a 10}{\log_a 3}= \frac{\frac{\log_a 10}{\log_a b}}{\frac{\log_a 3}{\log_a b}} = \frac{\log_b 10}{\log_b 3}$$ so the choice of base doesn't affect the result. $\endgroup$
    – MPW
    Sep 1 '16 at 17:07
  • $\begingroup$ @MPW Since $a$ is completely arbitrary (provided $a>0$ and $a\ne1$), there's no need to do that computation. $\endgroup$
    – egreg
    Sep 1 '16 at 17:12
  • $\begingroup$ I know you know that. I just meant to demonstrate to OP explicitly that they are the same. $\endgroup$
    – MPW
    Sep 1 '16 at 18:02
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Possibly the best way to see this is to write a change to an arbitrary base instead of choosing either $\log_{10}$ or $\ln$ initially. (There is already an answer that does that.) But I think it's also noteworthy that the change-of-base formula itself provides an easy derivation of the fact that $$\frac{\log10}{\log3}=\frac{\ln10}{\ln3}.$$

Starting with $\frac{\ln10}{\ln3}$ (for example), simply apply the formula to change the base from $e$ to $10$ on both the numerator and denominator: \begin{align} \ln10 &= \log_e 10 = \frac{\log_{10} 10}{\log_{10} e}, \\ \ln3 &= \log_e 3 = \frac{\log_{10} 3}{\log_{10} e}, \\ \frac{\ln10}{\ln3} &= \frac{\left( \frac{\log_{10} 10}{\log_{10} e} \right)} {\left(\frac{\log_{10} 3}{\log_{10} e} \right)} \end{align} Multiply both the numerator and denominator of the right-hand side of the last equation by $log_{10} e$, and we find that $$\frac{\ln10}{\ln3} = \frac{\log_{10} 10}{\log_{10} 3} $$

In other words, every time we change the base of both the numerator and denominator from base $b$ to base $c$ simultaneously, we divide both the numerator and denominator by $\log_c b$, and those two operations cancel each other out.

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In your question note that the definition of the logarithm implies that $$\log_310=x \iff 3^x=10 \tag 1.$$ In general $$\log_bw=x \iff b ^x=w \tag 2$$ where $b>0, b\neq 1$.

Now take the logarithm of both sides to some base $c>0$, $c\neq 1$ and use the fact that $\log_c(b^x)=x\log_cb $ to get that $$x\log_cb=\log_cw\implies x=\frac{\log_cw}{\log_cb}.\tag 3$$ Equations $(2)$ and $(3)$ imply that$$\log_bw=\frac{\log_cw}{\log_cb}. \tag 4$$ Take $c=10$ and $c=e$ in equation $(4)$ to get the equality $$\log_{3}10=\frac{\log_{10}10}{\log_{10}3}=\frac{\ln10}{\ln3}. \tag 5$$

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