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I'm trying to reboot my memory with the long-forgotten algebra that I used to know — way back in high school.

Please, help put my memory back on track? What are the algebraic steps that are missing from the transformation of this algebraic expression?

\begin{align*} 2^{\log_2 n+1} - 1 &= \quad?\\ &= \quad?\\ &= \quad?\\ &= \quad?\\ &= \quad...\\ &= \quad2n - 1\\ \end{align*}

Me being out of high school for a couple years at this point — and my algebra chops being practically non-existent, as a result — I would not be offended in the least if you'd elaborate on your answer, at the ELI5-level.

Like, I recall the meanings of one or two mathematical concepts like, "Multiplicative Property", "Commutative Property", and what-have-you.

I've already given it my best shot (like, 4 days of trawling Google and YouTube). I've proven to myself that plugging any arbitrary value into $n$ on both sides of the equation, works out as expected.

But I'm stumped when I try to apply what little algebra I do recall, to how exactly to end up with the $2n - 1$ expression. Help a guy out?

Thanks in advance.

EDIT: Oops! I meant to say, $2^{\log_2 n+1} - 1$ instead of $2^{\log_2 n} - 1$ My bad. I've corrected my innocent typo to make the $n+1$ exponent, look exactly like it's typeset in the book from which it originates. (see page 31 | (Equation 2.8))

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    It is $n-1$. There are no computing steps, just the *definition* of logarithm. For a natural number $m$, if $n=2^m$ then, by definition, we say that $m=\log_2 n$, so that $2^{\log_2 n}=n$. – 2017-02-02
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    Thanks @MiguelAtencia. Actually, I meant to say, $2^{\log_2 (n+1)} - 1$ instead of $2^{\log_2 (n)} - 1$ My bad. – 2017-02-02
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    Are you sure you mean $2^{\log_2(n+1)}$ rather than $2^{\log_2(n) + 1}$? – 2017-02-02
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    Like I said in a comment below @Hurkyl Any typos are innocent mistrakes due to the inherent awkwardness of manually translating to MathJax, from what I'm reading on [_page 31 (Equation 2.8) of the book from where I'm quoting the equation_](https://people.cs.vt.edu/shaffer/Book/Java3e20100119.pdf). – 2017-02-02

2 Answers 2

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In general, for real $a, b\in \mathbb R, \;b\gt0$: $$\quad a^{\large \log_a(b)} = b.$$

In keeping with this, your problem evaluates to:

$$2^{\large\log_2(n+1)} - 1 = n+1-1=n \neq 2n-1$$

In the event that ou are trying to evaluate: $2^{\log_2 n + 1} - 1 = 2^{ \large\log_2(n) + 1} - 1$, where the argument of $\log_2$ is strictly $n$, then, we have

$$\left(2^{\large\log_2(n) + 1}\right) - 1 = \left(\underbrace{2^{\large\log_2(n)}}_n\times 2^1\right) - 1 = 2n-1$$

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    Thanks @amWhy. Actually, I meant to say, $2^{\log_2 (n+1)} - 1$ instead of $2^{\log_2 (n)} - 1$ My bad. But you have definitely swept at least some of the spider webs from the disused attic that is the mathematical hemisphere of my brain. I will see if I can get to $2n - 1$ using your nifty formula. Thanks again. – 2017-02-02
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    I'm not "_changing the question in order to find some expression_...". It was an innocent mistake caused by me manually transcribing to the cumbersome LaTeX notation, from what I actually am reading in a text book. Don't believe me? [_See page 31 | Equation (2.3)_](https://people.cs.vt.edu/shaffer/Book/Java3e20100119.pdf). – 2017-02-02
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    Yes, I believe you. I'll remove my comment; I'm sorry if it upset you. – 2017-02-02
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    I also included, after answering your question, the question that may the text may have intended (even texts sometimes have errors in printing!) . At any rate, have a look. – 2017-02-02
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    Awesome! Thanks! Incidentally, it's [_(Equation 2.8) in the book_](https://people.cs.vt.edu/shaffer/Book/Java3e20100119.pdf). Not _(2.3)_. I can see how the $n - 1$ on the R.H. side, comes from applying your logarithm formula to the L.H. side. Now. I just need to stare at your answer for a few hours in order for the rest to sink in. You could fast-track the learning process for me though, if you'd just remind me why we need to multiply by $2^1$ on the L.H. side? I'm not _grokking_ why we need that. Something to do with logarithms again? BTW, I dropped out the semester before logarithms :¬) – 2017-02-02
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    Remember that $a^{b+c} = a^b\cdot a^c$....Similarly, $$(2^{\large \log_2(n) + 1}) -1= (2^{\large \log_2(n)}\cdot 2^1) - 1 =( n\cdot 2 )-1 = 2n-1.$$ – 2017-02-02
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    Just checked your link to the book. In the book, the author's statement of the problem is somewhat ambiguous, so I can totally understand the confusion. What I see there is $$2^{\large \log_2n + 1} -1$$ which is entirely clear that $\log_2 n + 1$ means $\log_2(n)+ 1$ (which in this case it does). I hope I've cleared things up in my answer and comments. – 2017-02-02
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    Ah yes! Now I remember! The old, $a^{b+c} = a^b \cdot a^c$ Of course. You're a mathematical genius @amWhy! Thanks! I tried to select your answer. But I don't have enough "_rep_". Some elitist douche from [_my first post_](https://math.stackexchange.com/questions/2111057/why-is-the-manual-summation-of-an-n-log-n-equation-not-equal-to-the-programmatic) is cyberstalking me, and downvoted me because I made them look bad by answering a question they didn't know the answer to. Plus, my earlier comments to you were deleted. So thanks again. – 2017-02-02
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Your answer is wrong! $$ a^{log_cb}=b^{log_ca} $$ so $$ 2^{log_2(n+1)} = (n+1)^{log_22} = (n+1)^1 = n+1 $$ means your answer is $ n $.

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    he has typo $ 2^{log_2(n+1)} $ instead of $ 2^{log_2(n) + 1} $. question have to be edited – 2017-02-02
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    I have wrote answer as $n$ too in last line, "means your answer in $n$". – 2017-02-02
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    I did not see your last line. Apologies if I misunderstood. – 2017-02-02
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    No need to apologies :) – 2017-02-02
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    Like I said in a comment above @AliMajedHA Any typos are innocent mistrakes due to the inherent awkwardness of manually translating to MathJax, from what I'm reading on [_page 31 (Equation 2.8) of the book from where I'm quoting the equation_](https://people.cs.vt.edu/shaffer/Book/Java3e20100119.pdf). – 2017-02-02