# What are the algebraic steps that are missing from the evaluation of this algebraic expression?

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))

• 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 $2^{\log_2 n}=n$. – Miguel Feb 2 '17 at 18:14
• Thanks @MiguelAtencia. Actually, I meant to say, $2^{\log_2 (n+1)} - 1$ instead of $2^{\log_2 (n)} - 1$ My bad. – algoHolic Feb 2 '17 at 18:31
• Are you sure you mean $2^{\log_2(n+1)}$ rather than $2^{\log_2(n) + 1}$? – Hurkyl Feb 2 '17 at 18:32
• 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. – algoHolic Feb 2 '17 at 20:38

## 2 Answers

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

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

• 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. – algoHolic Feb 2 '17 at 18:30
• 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). – algoHolic Feb 2 '17 at 18:46
• Yes, I believe you. I'll remove my comment; I'm sorry if it upset you. – Namaste Feb 2 '17 at 18:49
• 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. – Namaste Feb 2 '17 at 18:51
• Awesome! Thanks! Incidentally, it's (Equation 2.8) in the book. 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 :¬) – algoHolic Feb 2 '17 at 19:09

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 $n$.

• he has typo $2^{log_2(n+1)}$ instead of $2^{log_2(n) + 1}$. question have to be edited – user411775 Feb 2 '17 at 18:49
• I have wrote answer as $n$ too in last line, "means your answer in $n$". – user411775 Feb 2 '17 at 18:55
• I did not see your last line. Apologies if I misunderstood. – Namaste Feb 2 '17 at 18:56
• No need to apologies :) – user411775 Feb 2 '17 at 18:57
• 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. – algoHolic Feb 2 '17 at 20:42