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This is a question on manipulating polynomials from Udacity's Free Course in collaboration with Georgia Tech, Introduction to Graduate Algorithms, video number 190 (videos are downloadable for free), in the Divide and Conquer section.

$4^{\log_2n} = n^2$

$3^{\log_2n} = n^c$

$3 = 2^{log_23}$

How does one get $3 = 2^{\log_23}$?

I looked at the rules here: https://www.rapidtables.com/math/algebra/Logarithm.html
and tried to understand the rule by setting for example $3 = \log_3(9^1) = 1 \cdot \log_3(9)$ but cannot extrapolate to the original equation at the top.

mLstudent33
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    Welcome to MSE. Please type your questions instead of posting images. Images can't be browsed and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial – saulspatz Feb 02 '21 at 02:50

4 Answers4

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  1. $3 = 2^{log_2^3}$
  2. Let $\log_2 3= \lambda$

Thus, the equation becomes: $$3=2^{log_2^3}=2^\lambda$$

But, from equation 2, we can see that $\log_2 3=\lambda$. Or, $2^\lambda= 3$ Hope this answers your question.

Smriti Sivakumar
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For any base $a>0,a\ne 1$, we know that $a^x$ and $\log _ax$ are inverse functions. If we let $f(x)=a^x$ and $f^{-1}(x)=\log _ax$, we can easily say that $$f(f^{-1}(x))=a^{\log _ax}=x$$ In this case if $a=2$ and $x=3$, we have $$3=2^{\log _23}$$

5201314
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I am not sure if you are misreading the expression, or just mistyped it, but the exponent on the $2$ is not $\log_2^3$ (which doesn't mean anything) but rather $\log_2 3$. As for why $2 = 2^{\log_2 3}$, this is a case of a general identity:

$$b^{\log_b x} = x$$ for any $x>0$ and any base $b>0$.

As for why this identity is true: The function $f(x) = \log_b x$ and $g(x) = b^x$ are inverse functions. (Indeed, that's usually how the logarithm function is defined, as the inverse of the exponential function.) That means that $f(g(x)) = x$ and $g(f(x))=x$. The first one of these equations says exactly that $b^{\log_b x} = x$.

Here's an informal way of explaining it that might help you understand better:

The expression $\log_2 3$ can be thought of as the answer to the question "What power do you raise $2$ to, if you want to get $3$?" The expression $2^{\log_2 3}$ means "raise 2 to the power that, if you raised $2$ to it, would produce $3$." When you say it that way, it should be obvious that the result is $3$.

If it's still confusing, try it again with different numbers, like $10$ and $1000$. Then $\log_{10} 1000$ means "find the exponent which, when $10$ is raised to it, produces $1000$." The answer to this is $3$. Now $10^{\log_{10} 1000}$ means "Raise $10$ to the $3$rd power", which produces $1000$, of course. So $10^{\log_{10} 1000} = 1000$.

mweiss
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  • I appreciate your use of a more obvious example of 10 and 1000. I was just about to edit as I was writing on scrap paper how $x = b^y$ and $y = log_b(x)$ are inverses? Then I let $x = 4^(log_2(n)) $ which has to be the inverse of $log_2(n) = log_4(x)$. Now I had the situation where $2^? = n$ and $4^? = x$ and subbed in 2 which means $2^2 = n$ and $4^2 = x$ and to get from n to x I have to square n. – mLstudent33 Feb 02 '21 at 03:36
  • I am sorry for that superscript error in my comment, still getting the hang of this. – mLstudent33 Feb 02 '21 at 03:39
  • $x = 4^{log_2(n)} $ – mLstudent33 Feb 02 '21 at 03:45
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A logarithm is just an exponent. It’s just the exponent that the base needs to carry in order to produce the given value.

So if you think about $3^?=9$, you know $2$ is the exponent that $3$ needs to carry to produce $9$. That means $\log_3 9= 2$.

So now if you consider $2^?=3$, you are asking what exponent $2$ needs to carry to produce $3$. This is exactly what $\log_2 3$ is, so of course it is true that $2^{\log_2 3}=3$.

Likewise $5^{\log_5 3}=3$ because $\log_5 3$ is the exponent that $5$ must carry to produce $3$.

And $17^{\log_{17}91}=91$ because $\log_{17} 91$ is the exponent that $17$ must carry to produce $91$.

And so on.

MPW
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