My Math Forum Confused with indices

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 January 9th, 2015, 02:03 PM #1 Newbie   Joined: Oct 2014 From: england Posts: 23 Thanks: 0 Confused with indices Hi, so recently I've been learning about powers and indices. Were currently learning how to work out algebraic powers. I was given this question... 3y^2+3 = 9^2y I'm confused as to how the +3 has become a power in the first place? e.g would it have been 3y^2 * 27 = 9^2y and then 3y^2 * 3^3 = 9^2y ? Thanks for any advice. Last edited by skipjack; January 9th, 2015 at 03:45 PM.
January 9th, 2015, 02:37 PM   #2
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Quote:
 3y^2+3 = 9^2y
As posted, I see $\displaystyle 3y^2 + 3 = 9^2y$ or $\displaystyle 3y^2+3 =81y$

Is it possible you meant $\displaystyle 3y^2+3 = 9^{2y}$ ?

Maybe something else?

Recall your order of operations and use some grouping symbols ... parentheses would be nice ... to clarify your equation.

Thanks.

Last edited by skipjack; January 9th, 2015 at 03:45 PM.

January 9th, 2015, 03:29 PM   #3
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Hello, nicevans1!

Some parentheses would be welcome.
I'll guess at what you meant.

Quote:
 $3^{y^2+3} \:=\: 9^{2y}$

We have: $\:3^{y^2+3} \:=\:9^{2y} \quad\Rightarrow\quad 3^{y^2+3} \:=\:(3^2)^{2y} \quad\Rightarrow\quad 3^{y^2+3} \:=\:3^{4y}$

Hence: $\:y^2+3 \:=\:4y \quad\Rightarrow\quad y^2 - 4y + 3 \:=\:0 \quad\Rightarrow\quad (y-1)(y-3) \:=\:0$

Therefore: $\:y \:=\: 1,\,3$

January 13th, 2015, 11:04 AM   #4
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Quote:
 Originally Posted by skeeter Recall your order of operations and use some grouping symbols ... parentheses would be nice ... to clarify your equation. Thanks.
Sorry, I was rushing

Quote:
 Originally Posted by soroban Hello, nicevans1! We have: $\displaystyle \:3^{y^2+3} \:=\:9^{2y} \quad\Rightarrow\quad 3^{y^2+3} \:=\:(3^2)^{2y} \quad\Rightarrow\quad 3^{y^2+3} \:=\:3^{4y}$ Hence: $\displaystyle \:y^2+3 \:=\:4y \quad\Rightarrow\quad y^2 - 4y + 3 \:=\:0 \quad\Rightarrow\quad (y-1)(y-3) \:=\:0$ Therefore: $\displaystyle \:y \:=\: 1,\,3$
Cool That's exactly how it is in the book....

So first can you explain (as this question really took me up the wrong alley)

What is actually an indice and what is not?

Is the first Y a power and if not why is it the same size in text as the y at end? Also why is that 2 smaller than the 3 in terms of size of text? The 2 is also raised higher than the rest of the powers, why??

I honestly presumed it was a mis print in the book and its really thrown me.

Last edited by greg1313; January 13th, 2015 at 11:17 AM.

 January 13th, 2015, 02:14 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 In the United States, an "index" is simply a subscript or superscript labeling a variable or constant. But our English friends use it to denote what we would call an "exponent". In the very simplest case, when it is a positive integer, it just tells how many times the base is multiplied by itself: $a^2= a*a$, $a^3= a*a*a$, $a^5= a*a*a*a*a$, etc. From that you can, perhaps by induction on n and m, that $(a^n)(a^m)= a^{n+ m}$ and $(a^n)^m= a^{nm}$. Those are nice properties so we extend to other numbers. If, for example, m= 0, we have $(a^0)(a^m)= a^{0+ m}= a^m$. In order to have $a^0a^m= a^m$ we must define $a^0= 1$. Now, to extend to negative integers, look at $(a^n)(a^{-n})= a^{n- n}= a^0= 1$. We must define $a^{-n}= \frac{1}{a^n}$ for n> 0. Next, look at m= 1/n. $(a^n)^{1/n}= a^{\frac{n}{n}}= a^1= 1$. In order for the rule $(a^m)^n= a^{mn}$ for exponents of the form $\frac{1}{n}$, then we must define $a^{\frac{1}{n}$ to be $\sqrt[n]{a}$. To continue to all real numbers, we have to switch from 'algebraic' to 'analytic' properties. A function is said to be "continuous at x= a" if and only if $\{x\}$ is a sequence of numbers that converge to a, then the sequence $\{f(x_m)\}$ converges to f(a). (That is one of many equivalent definitions of "continuous".) If x is any real number, then there exist a sequence of rational numbers, $\{x_n\}$ that converges to x so, in order to have $f(x)= a^x$ be continuous, we define $\mathtype a^x= \lim_{n\to \infty} f(x_n)$ where $\{x_n\}$ is any sequence of rational numbers that converges to x. Thanks from topsquark
 January 13th, 2015, 03:27 PM #6 Banned Camp   Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 Country Boy, Don't try to use $and$. -------------------------------------------------------------------------------- They don't work. Instead use [ MATH ], [ /MATH ], but take out the spaces between "[" and "M" and "H" and "]." Also, take out the spaces between "[" and "/" and "H" and "]" again.
January 13th, 2015, 04:42 PM   #7
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You don't need any of Country Boy's definitions if you start from the other end. We want(ed) a function such that $f(xy) = f(x) + f(y)$ to make multiplication and division easier. It turns out that the only non-trivial functions that do this are of the form $$L(x) = c\int_1^{|x|} \tfrac1t \, \mathrm d t \qquad \text{where x \ne 0}$$
So, if we define the natural logarithm to be $$\log x = \int_1^x \tfrac1t \, \mathrm d t \qquad \text{where x \gt 0}$$and define the number $\mathrm e$ to be the (unique) solution to $\log x = 1$, we can then define $\mathrm e^x$ to be the inverse function of $\log x$. That is$$y = \log x \implies \mathrm e^y = x$$

This defines $\mathrm e^x$ for all (real) $x$ without any messing around. It also gets us $a^x = \mathrm e^{x \log a}$ for any positive $a$, and all the nice properties that we with exponentiation to have can be proved.

We get logarithms in other bases either as inverse functions to $a^x$ so that $$a^x = y \implies \log_a y = x$$or using the logarithmic identity$$\log_a x = {\log x \over \log a }$$

I quote Country Boy's post in its entirety with tags corrected.
Quote:
 Originally Posted by Country Boy In the United States, an "index" is simply a subscript or superscript labelling a variable or constant. But our English friends use it to denote what we would call an "exponent". In the very simplest case, when it is a positive integer, it just tells how many times the base is multiplied by itself: $\displaystyle a^2= a*a$, $\displaystyle a^3= a*a*a$, $\displaystyle a^5= a*a*a*a*a$, etc. From that you can, perhaps by induction on n and m, that $\displaystyle (a^n)(a^m)= a^{n+ m}$ and $\displaystyle (a^n)^m= a^{nm}$. Those are nice properties so we extend to other numbers. If, for example, m= 0, we have $\displaystyle (a^0)(a^m)= a^{0+ m}= a^m$. In order to have $\displaystyle a^0a^m= a^m$ we must define $\displaystyle a^0= 1$. Now, to extend to negative integers, look at $\displaystyle (a^n)(a^{-n})= a^{n- n}= a^0= 1$. We must define $\displaystyle a^{-n}= \frac{1}{a^n}$ for n> 0. Next, look at m= 1/n. $\displaystyle (a^n)^{1/n}= a^{\frac{n}{n}}= a^1= 1$. In order for the rule $\displaystyle (a^m)^n= a^{mn}$ for exponents of the form $\displaystyle \frac{1}{n}$, then we must define $\displaystyle a^{\frac{1}{n}}$ to be $\displaystyle \sqrt[n]{a}$. To continue to all real numbers, we have to switch from 'algebraic' to 'analytic' properties. A function is said to be "continuous at x= a" if and only if $\displaystyle \{x\}$ is a sequence of numbers that converge to a, then the sequence $\displaystyle \{f(x_m)\}$ converges to f(a). (That is one of many equivalent definitions of "continuous".) If x is any real number, then there exist a sequence of rational numbers, $\displaystyle \{x_n\}$ that converges to x so, in order to have $\displaystyle f(x)= a^x$ be continuous, we define $\displaystyle a^x= \lim_{n\to \infty} f(x_n)$ where $\displaystyle \{x_n\}$ is any sequence of rational numbers that converges to x.

Last edited by v8archie; January 13th, 2015 at 05:07 PM.

January 14th, 2015, 01:59 PM   #8
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Quote:
 Originally Posted by Math Message Board tutor Country Boy, Don't try to use $and$. -------------------------------------------------------------------------------- They don't work. Instead use [ MATH ], [ /MATH ], but take out the spaces between "[" and "M" and "H" and "]." Also, take out the spaces between "[" and "/" and "H" and "]" again.
Thank you!

January 18th, 2015, 01:22 PM   #9
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Quote:
 Originally Posted by nicevans1 What is actually an indice and what is not? Is the first Y a power and if not why is it the same size in text as the y at end? Also why is that 2 smaller than the 3 in terms of size of text? The 2 is also raised higher than the rest of the powers, why?? I honestly presumed it was a mis print in the book
Sorry about this and thanks for the extensive replies, but I did not word my question correctly!!

When I said what is an indice, I meant in terms of the question I posted about.

What is an exponent in the question as all the numbers and letters seem a different levels? So im confused as to what is an exponent and what is not!

3y^2+3=9^2y

Thankyou

 January 19th, 2015, 08:33 AM #10 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित you need to use latex to write the expression in appropriate form like you write in a paper. If you don't know latex you can use brackets ( ) for understanding. like $\displaystyle 3^{y^2+3}=9^{2y}$ can be written as 3^(y^2+3)=9^(2y) The exponent of a number says how many times to use that number in a multiplication, its written at the top of the number or after the symbol ^ if only one constant or variable is used as exponent, brackets ( ) are not required. If more than one constant or variable is used as exponent we need brackets ( ) for clear understanding.

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