I've recently started learning about logarithms and was just curious as to why the base "b" in the "change the base" formula doesn't matter, as in it doesn't matter what value you put in for it, it still yields the same answer. Also, why when taking the logarithm of both sides of an equation it also doesn't matter the base of the logarithm, you still get the same answer for x?

enter image description here Such as in this question "a" can be any number and you still get the same answer of 45

enter image description here

And, again, it doesn't matter the value of "b", it gives you the same answer.

If someone could explain, I'd be very grateful.

Thanks a lot!

  • $\begingroup$ For the first one, did you try combining the logarithms using the rules to see why the base won't matter? $\endgroup$
    – Triatticus
    Jun 26 '21 at 21:32
  • $\begingroup$ It doesn't matter only in situations, where constant factor is not important, in other situations it matter. $\endgroup$
    – zkutch
    Jun 26 '21 at 21:35
  • $\begingroup$ Not a proof, but I find it helpful to think about changing the bases again in your formula $$\log_a(x) = \frac{\log_b(x)}{\log_b(a)} = \frac{\frac{\log_c(x)}{\log_c(b)}}{\frac{\log_c(a)}{\log_c(b)}}$$ And then the $\frac{1}{\log_c(b)}$ in the numerator and denominator cancel each other out. $\endgroup$
    – TomKern
    Jun 26 '21 at 22:45
  • $\begingroup$ a can be any positive number other than 1. $\endgroup$ Jun 27 '21 at 7:42

The base does not matter because we can prove that $$ \log_a(x)=\frac{\log_b(x)}{\log_b(a)} \, , $$ for any positive $b$ such that $b\neq1$. To do this, let $y=\log_a(x)$. By definition, this means that $a^y=x$. Hence, \begin{align} \log_b(a^y)&=\log_b(x) \\[5pt] y\log_b(a) &= \log_b(x) \\[5pt] y = \log_a(x)&=\frac{\log_b(x)}{\log_b(a)} \, . \end{align} Notice that at no point in this derivation did it matter what base $\log_b$ is (the only requirement was that the function $\log_b$ actually made sense—the base couldn't be negative or equal to $1$). The power rule $\log_b(a^y)=y\log_b(a)$ is valid for any positive $b$ such that $b\neq1$, and so the same is true of the change of base formula.


The base doesn't matter because this property is based on the fact that

$$\log(a^c)=c\log a$$

So when you have, based on another property of logarithms,

$$a^{\log_a x}=x$$

You can take a logarithm on both sides with any base you want${}^{\star}$ to get rid of that exponent. Say you choose a logarithm to the base $b$:

$$\log_b(a^{\log_a x})=\log_b x$$

$$\log_a x \times \log_b a = \log_b x$$

$$\log_a x = \frac{\log_b x}{\log_b a}$$

${}^{\star}$ Of course, there are still some numbers that you can't take as bases in any logarithm, irrespective of which property you use: negative numbers, zero, and one.


One way to think about $\ \log_c d = z\ $ is that it is simply a different, yet equivalent way of writing $\ c^z=d.$

Let $\ a>0,\ x>0.\ $ Define $\ y_1=\log_a x.\ $

Now, for any $\ b>0,\ $ all of the following holds:

$$\text{Let}\ y_2=\log_b a.\ \text{Then, from our equivalent way of writing,}$$

$$b^{y_2}=a\quad \text{and}\quad a^{y_1}=x.\ \text{So,}$$

$$\color{red}{x}=a^{y_1}=\left(b^{y_2}\right)^{y_1}=\color{red}{b^{y_2 y_1}}.\ \text{Therefore,}$$

$$ \log_b x = y_2 y_1\quad \text{using again our equivalent way of writing.}$$

$$ \log_b x = y_2 y_1= \log_b a\ \cdot\ \log_a x.\quad \text{Rearranging, we get}$$

$$ \log_a x = \frac{\log_b x} {\log_b a}.$$ $$$$ We can also apply the methods used above to your first equation, or we can just bypass all this stuff and go straight to the laws of logarithms. Let $\ y = \log_a x.\ $ Then,

$y=\log_a x = 2\log_a 3 + \log_a 5=\log_a 3^2 + \log_a 5 = \log_a 9 + \log_a 5=\log_a (9\times 5) = \log_a 45.$

But using our equivalent way of writing, we have: $\ x = a^y = 45.$


As we have

$$\log_bf(x)=\log_af(x) \cdot \log_b a $$

Then, equation

$$\log_a x = 2 \log_a 3 + \log_a 3 \quad (1)$$

for suitable $b$, can be multiplied on both sides by $\log_b a$ $$\log_b a \cdot \log_a x = \log_b a \cdot (2 \log_a 3 + \log_a 3) $$

which gives

$$\log_b x = 2 \log_b 3 + \log_b 3 \quad (2)$$

So, for suitable $b$, equations $(1)$ and $(2)$ has same solutions.


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