Solving Equation

pellegrind · #1 Sep 17, 2009, 07:04 PM

my daugheter has an algebra problem that she does nto know how to do

It is 5 to the 4th power over 5 to the 8th power

Thank you

Ginny Finny · Sep 17, 2009, 09:00 PM

5 divided by5 =1. you are now left with 1 to the power of 4 divided by1 to the power of 8. 4 divided by 8 can be simplified into 1 divided by two. you now hav 1 raised to the power of 1 divided by 1 raised to the power of two. no matter what power you put next to 1, it still = 1. so you are now left with 1 divided by 1 which = 1.

ebaines · Sep 18, 2009, 06:13 AM

The problem is this:

$ \frac {5^4} {5^8} $

This can be apprpoached in a number of ways - I'll start with the most fundamental and then show how to do it really quickly.

First, recall that the term "5 to the fourth power" means 5 multipled by itself 4 times. Similarly, 5 to the 8th power is 5 multiplied by itself 8 times. So this expression is the same as:

$ \frac {5*5*5*5} {5*5*5*5*5*5*5*5} $

Since $\frac 5 5 = 1$ , can eliminate a matching set of 5's from both the numerator and denominator, and do this 4 times:

$ \frac {5*5*5*5} {5*5*5*5*5*5*5*5} = \frac 1 {5*5*5*5} = \frac 1 {5^4} = \frac 1 {625} $

The short hand way to do this is to realize that putting 5 to the 8th power in the denominator is like multiplying by 5 to the minus eighth power. So:

$ \frac {5^4} {5^8} = 5^4 * 5^{-8} $

Now, whenever you are multiplying a number raised to a power by that same number raised to another power, it's equivalent to that number raised to the sum of the two powers. That is:

$ a^b * a^ c = a^{(b+c)} $

Similarly, when you are dividing a number raised to a power by that same number raised to another power, it's equivalent to that number raised to the difference of the two powers. That is:

$ \frac {a^b} { a^ c} = a^{(b-c)} $

So for this problem:

$ \frac {5^4} {5^8} = 5^4 * 5^ {-8} = 5^ {(4-8)} = 5^ {-4} = \frac 1 {5^4} = \frac 1 {625} $

If you can remember the rules for adding or subtracting exponents, this type of problem can be solved in seconds.

Ginny Finny · Sep 18, 2009, 04:24 PM

you may like to explain that to my math teacher then. depending on what rules you are supposed to use, you can get different answers. eg, inserting logs.

Unknown008 · Sep 19, 2009, 01:18 AM

Quote:

Originally Posted by Ginny Finny

you may like to explain that to my math teacher then. depending on what rules you are supposed to use, you can get different answers. eg, inserting logs.

If you type those in a calculator, you'll get the answer that ebaines already gave.

Ok, I'll use other examples.

$\frac{2^3}{2^2}$

$2^3$ is 8, whereas $2^2$ is 4. And 8/4 gives 2, not 1.

$\frac{3^3}{3^4}$

$3^3$ is 27, whereas $3^4$ is 814. And 27/81 gives 1/3, not 1.

Using logs? Ok, let's do it.

$\frac{2^3}{2^2}=x$

Let x be the answer.

$2^3=2^2x$

Insert log;

$log(2^3)=log(2^2x)$

Separate log(2^2x) and remove the power from log(2^3)

$3log(2)=log(2^2) + log(x)$

Remove power from log(2^2)

$3log(2)=2log(2) + log(x)$

$3log(2)-2log(2) = log(x)$

Let log(2) = y

$3y-2y = log(x)$

$y = log(x)$

Substitute y by log(2)

$log(2) = log(x)$

So, x = 2.

Perito · Sep 19, 2009, 05:28 AM

Quote:

you may like to explain that to my math teacher then. depending on what rules you are supposed to use, you can get different answers. eg, inserting logs.

You will not get different answers if you do it correctly, no matter how you do the problem.



At Ask Me Help Desk you can ask questions in any topic and have them answered for free by our experts. To ask questions or participate in answering them you must register for a free account. By registering you will be able to: Get free answers from experts in any of our 300+ topics. Accept money for answers that you provide. Communicate privately with other members (PM). See fewer ads. Sign Up!